OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Exercise 8 (A)

Question 1. Differentiate with respect to x :
(i) \( x^5 \)
(ii) \( 6x^8 \)
(iii) \( x^{3/4} \)
(iv) \( 4\sqrt{x} \)
(v) \( 8x^{-3/4} \)
(vi) \( \sqrt{x^3} \)
(vii) \( \frac{9}{x} \)
(viii) \( \frac{7}{x^2} \)
Answer:
(i) Let \( y = x^5 \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx} (x^5) \)
\( \implies \frac{dy}{dx} = 5x^{5-1} \)
\( \implies \frac{dy}{dx} = 5x^4 \).
(ii) Let \( y = 6x^8 \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx} (6x^8) \)
\( \implies \frac{dy}{dx} = 6 \frac{d}{dx} (x^8) \)
\( \implies \frac{dy}{dx} = 6 \times 8x^{8-1} \)
\( \implies \frac{dy}{dx} = 48x^7 \).
(iii) Let \( y = x^{3/4} \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx} (x^{3/4}) \)
\( \implies \frac{dy}{dx} = \frac{3}{4} x^{\frac{3}{4}-1} \)
\( \implies \frac{dy}{dx} = \frac{3}{4} x^{-\frac{1}{4}} \)
(iv) Let \( y = 4\sqrt{x} \).
We can write \( \sqrt{x} \) as \( x^{1/2} \). So, \( y = 4x^{1/2} \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx} (4x^{1/2}) \)
\( \implies \frac{dy}{dx} = 4 \frac{d}{dx} (x^{1/2}) \)
\( \implies \frac{dy}{dx} = 4 \times \frac{1}{2} x^{\frac{1}{2}-1} \)
\( \implies \frac{dy}{dx} = 2x^{-\frac{1}{2}} \)
\( \implies \frac{dy}{dx} = \frac{2}{\sqrt{x}} \).
(v) Let \( y = 8x^{-3/4} \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx} (8x^{-3/4}) \)
\( \implies \frac{dy}{dx} = 8 \frac{d}{dx} (x^{-3/4}) \)
\( \implies \frac{dy}{dx} = 8 \times \left(-\frac{3}{4}\right) x^{-\frac{3}{4}-1} \)
\( \implies \frac{dy}{dx} = -6 x^{-\frac{7}{4}} \).
(vi) Let \( y = \sqrt{x^3} \).
We can write \( \sqrt{x^3} \) as \( x^{3/2} \). So, \( y = x^{3/2} \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx} (x^{3/2}) \)
\( \implies \frac{dy}{dx} = \frac{3}{2} x^{\frac{3}{2}-1} \)
\( \implies \frac{dy}{dx} = \frac{3}{2} x^{\frac{1}{2}} \)
\( \implies \frac{dy}{dx} = \frac{3}{2} \sqrt{x} \).
(vii) Let \( y = \frac{9}{x} \).
We can write \( \frac{9}{x} \) as \( 9x^{-1} \). So, \( y = 9x^{-1} \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx} (9x^{-1}) \)
\( \implies \frac{dy}{dx} = 9 \frac{d}{dx} (x^{-1}) \)
\( \implies \frac{dy}{dx} = 9 \times (-1) x^{-1-1} \)
\( \implies \frac{dy}{dx} = -9x^{-2} \)
\( \implies \frac{dy}{dx} = \frac{-9}{x^2} \).
(viii) Let \( y = \frac{7}{x^2} \).
We can write \( \frac{7}{x^2} \) as \( 7x^{-2} \). So, \( y = 7x^{-2} \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx} (7x^{-2}) \)
\( \implies \frac{dy}{dx} = 7 \frac{d}{dx} (x^{-2}) \)
\( \implies \frac{dy}{dx} = 7 \times (-2) x^{-2-1} \)
\( \implies \frac{dy}{dx} = -14x^{-3} \)
\( \implies \frac{dy}{dx} = \frac{-14}{x^3} \).
In simple words: To find the derivative of \( x^n \), we multiply by 'n' and then subtract 1 from the power, making it \( nx^{n-1} \). If there is a constant number, like 6 or 4, it stays multiplied to the derivative. When x is in the bottom part of a fraction, like \( \frac{1}{x} \), we first write it as \( x^{-1} \) and then apply the same power rule.

🎯 Exam Tip: Remember the basic power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \). For expressions like \( \sqrt{x} \) or \( \frac{1}{x} \), convert them to power form first, like \( x^{1/2} \) and \( x^{-1} \), before differentiating. This makes the calculation easier and less prone to errors.

Question 2. Differentiate with respect to x :
(i) \( (2x + 3)^5 \)
(ii) \( (1 – x)^4 \)
(iii) \( \sqrt{8-7 x} \)
(iv) \( (3x^2 + 5)^9 \)
Answer:
(i) Let \( y = (2x + 3)^5 \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = 5(2x + 3)^{5-1} \times \frac{d}{dx}(2x + 3) \)
\( \implies \frac{dy}{dx} = 5(2x + 3)^4 \times (2 + 0) \)
\( \implies \frac{dy}{dx} = 10(2x + 3)^4 \).
(ii) Let \( y = (1 - x)^4 \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = 4(1 - x)^{4-1} \times \frac{d}{dx}(1 - x) \)
\( \implies \frac{dy}{dx} = 4(1 - x)^3 \times (0 - 1) \)
\( \implies \frac{dy}{dx} = -4(1 - x)^3 \).
(iii) Let \( y = \sqrt{8-7x} \).
We can write \( \sqrt{8-7x} \) as \( (8-7x)^{1/2} \). So, \( y = (8-7x)^{1/2} \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = \frac{1}{2}(8-7x)^{\frac{1}{2}-1} \times \frac{d}{dx}(8-7x) \)
\( \implies \frac{dy}{dx} = \frac{1}{2}(8-7x)^{-\frac{1}{2}} \times (0-7) \)
\( \implies \frac{dy}{dx} = \frac{-7}{2\sqrt{8-7x}} \).
(iv) Let \( y = (3x^2 + 5)^9 \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = 9(3x^2 + 5)^{9-1} \times \frac{d}{dx}(3x^2 + 5) \)
\( \implies \frac{dy}{dx} = 9(3x^2 + 5)^8 \times (3 \times 2x + 0) \)
\( \implies \frac{dy}{dx} = 9(3x^2 + 5)^8 \times 6x \)
\( \implies \frac{dy}{dx} = 54x(3x^2 + 5)^8 \).
In simple words: When you have a function inside another function (like something raised to a power), you use the chain rule. This means you differentiate the "outside" part first, keeping the "inside" part the same, and then you multiply that by the derivative of the "inside" part. This is like unpeeling an onion, layer by layer.

🎯 Exam Tip: Always remember to multiply by the derivative of the "inside" function when using the chain rule. A common mistake is to forget this final step, which leads to an incomplete answer.

Question 3. Differentiate with respect to x :
(i) \( \frac{2}{x}+\frac{1}{\sqrt{x}} \)
(ii) \( (2x – 1) (3x + 2) \)
(iii) \( x^4-2 x+\frac{1}{x^2} \)
(iv) \( 2x^2 (x + 1) + 2 \)
(v) \( \frac{3 x^4-x}{x^3} \)
Answer:
(i) Let \( y = \frac{2}{x}+\frac{1}{\sqrt{x}} \).
We can rewrite this as \( y = 2x^{-1} + x^{-1/2} \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx}(2x^{-1}) + \frac{d}{dx}(x^{-1/2}) \)
\( \implies \frac{dy}{dx} = 2(-1)x^{-1-1} + \left(-\frac{1}{2}\right)x^{-\frac{1}{2}-1} \)
\( \implies \frac{dy}{dx} = -2x^{-2} - \frac{1}{2}x^{-3/2} \)
\( \implies \frac{dy}{dx} = -\frac{2}{x^2} - \frac{1}{2x^{3/2}} \).
(ii) Let \( y = (2x - 1)(3x + 2) \).
First, expand the expression: \( y = 6x^2 + 4x - 3x - 2 \)
\( \implies y = 6x^2 + x - 2 \).
Now, we differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx}(6x^2) + \frac{d}{dx}(x) - \frac{d}{dx}(2) \)
\( \implies \frac{dy}{dx} = 6(2x) + 1 - 0 \)
\( \implies \frac{dy}{dx} = 12x + 1 \).
(iii) Let \( y = x^4 - 2x + \frac{1}{x^2} \).
We can rewrite this as \( y = x^4 - 2x + x^{-2} \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx}(x^4) - \frac{d}{dx}(2x) + \frac{d}{dx}(x^{-2}) \)
\( \implies \frac{dy}{dx} = 4x^3 - 2(1) + (-2)x^{-2-1} \)
\( \implies \frac{dy}{dx} = 4x^3 - 2 - 2x^{-3} \)
\( \implies \frac{dy}{dx} = 4x^3 - 2 - \frac{2}{x^3} \).
(iv) Let \( y = 2x^2(x + 1) + 2 \).
First, expand the expression: \( y = 2x^3 + 2x^2 + 2 \).
Now, we differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx}(2x^3) + \frac{d}{dx}(2x^2) + \frac{d}{dx}(2) \)
\( \implies \frac{dy}{dx} = 2(3x^2) + 2(2x) + 0 \)
\( \implies \frac{dy}{dx} = 6x^2 + 4x \).
(v) Let \( y = \frac{3x^4 - x}{x^3} \).
First, simplify the expression by dividing each term in the numerator by \( x^3 \).
\( y = \frac{3x^4}{x^3} - \frac{x}{x^3} \)
\( \implies y = 3x - x^{-2} \).
Now, we differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx}(3x) - \frac{d}{dx}(x^{-2}) \)
\( \implies \frac{dy}{dx} = 3(1) - (-2)x^{-2-1} \)
\( \implies \frac{dy}{dx} = 3 + 2x^{-3} \)
\( \implies \frac{dy}{dx} = 3 + \frac{2}{x^3} \).
In simple words: For problems with multiple terms, we can differentiate each term separately. If expressions are multiplied or divided, it's often easier to expand or simplify them first before taking the derivative. This helps to avoid using complex rules like the product or quotient rule if not strictly necessary.

🎯 Exam Tip: Always simplify algebraic expressions before differentiating whenever possible. This can turn a complicated product or quotient rule problem into a simpler sum/difference rule problem, reducing calculation errors.

Question 4. Differentiate with respect to x :
(i) \( \cos 7x \)
(ii) \( \tan ax \)
(iii) \( \sec 9x \)
(iv) \( \sin x^2 \)
(v) \( \cos \sqrt{x} \)
(vi) \( 2 \operatorname{cosec} bx^3 \)
Answer:
(i) Let \( y = \cos 7x \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = -\sin(7x) \times \frac{d}{dx}(7x) \)
\( \implies \frac{dy}{dx} = -\sin(7x) \times 7 \)
\( \implies \frac{dy}{dx} = -7\sin 7x \).
(ii) Let \( y = \tan ax \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = \sec^2(ax) \times \frac{d}{dx}(ax) \)
\( \implies \frac{dy}{dx} = \sec^2(ax) \times a \)
\( \implies \frac{dy}{dx} = a\sec^2 ax \).
(iii) Let \( y = \sec 9x \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = \sec(9x)\tan(9x) \times \frac{d}{dx}(9x) \)
\( \implies \frac{dy}{dx} = \sec(9x)\tan(9x) \times 9 \)
\( \implies \frac{dy}{dx} = 9\sec 9x \tan 9x \).
(iv) Let \( y = \sin x^2 \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = \cos(x^2) \times \frac{d}{dx}(x^2) \)
\( \implies \frac{dy}{dx} = \cos(x^2) \times 2x \)
\( \implies \frac{dy}{dx} = 2x \cos x^2 \).
(v) Let \( y = \cos \sqrt{x} \).
We can write \( \sqrt{x} \) as \( x^{1/2} \). So, \( y = \cos(x^{1/2}) \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = -\sin(x^{1/2}) \times \frac{d}{dx}(x^{1/2}) \)
\( \implies \frac{dy}{dx} = -\sin(\sqrt{x}) \times \frac{1}{2}x^{\frac{1}{2}-1} \)
\( \implies \frac{dy}{dx} = -\sin(\sqrt{x}) \times \frac{1}{2}x^{-\frac{1}{2}} \)
\( \implies \frac{dy}{dx} = -\frac{\sin \sqrt{x}}{2\sqrt{x}} \).
(vi) Let \( y = 2 \operatorname{cosec} bx^3 \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = 2 \left(-\operatorname{cosec}(bx^3)\cot(bx^3)\right) \times \frac{d}{dx}(bx^3) \)
\( \implies \frac{dy}{dx} = -2 \operatorname{cosec}(bx^3)\cot(bx^3) \times b(3x^2) \)
\( \implies \frac{dy}{dx} = -6bx^2 \operatorname{cosec}(bx^3)\cot(bx^3) \).
In simple words: When differentiating trigonometric functions, remember the standard derivative rules (e.g., derivative of cos is -sin). If the angle itself is a function of x (like 7x or \( x^2 \)), always multiply by the derivative of that angle using the chain rule. This means an extra step for these types of problems.

🎯 Exam Tip: Memorize the derivatives of basic trigonometric functions and practice applying the chain rule. Pay close attention to the sign changes, for example, \( \frac{d}{dx}(\cos x) = -\sin x \) and \( \frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x \).

Question 5. Differentiate with respect to x :
(i) \( \frac{x^3}{3 x-2} \)
(ii) \( \frac{x}{\sin x} \)
(iii) \( \frac{1+\cos x}{1-\cos x} \)
(iv) \( x^2-\sqrt{(1+x)} \)
(v) \( \sin 2x \cos^2 x \)
(vi) \( \tan^4 7x \)
(vii) \( \frac{1}{\sin x+\cos x} \)
(viii) \( \frac{1+\cos x}{x} \)
Answer:
(i) Let \( y = \frac{x^3}{3x-2} \).
We use the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \).
Here, \( u = x^3 \) and \( v = 3x-2 \).
\( \frac{du}{dx} = 3x^2 \) and \( \frac{dv}{dx} = 3 \).
\( \frac{dy}{dx} = \frac{(3x-2)(3x^2) - (x^3)(3)}{(3x-2)^2} \)
\( \implies \frac{dy}{dx} = \frac{9x^3 - 6x^2 - 3x^3}{(3x-2)^2} \)
\( \implies \frac{dy}{dx} = \frac{6x^3 - 6x^2}{(3x-2)^2} \)
\( \implies \frac{dy}{dx} = \frac{6x^2(x-1)}{(3x-2)^2} \).
(ii) Let \( y = \frac{x}{\sin x} \).
We use the quotient rule. Here, \( u = x \) and \( v = \sin x \).
\( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \cos x \).
\( \frac{dy}{dx} = \frac{(\sin x)(1) - (x)(\cos x)}{(\sin x)^2} \)
\( \implies \frac{dy}{dx} = \frac{\sin x - x\cos x}{\sin^2 x} \).
(iii) Let \( y = \frac{1+\cos x}{1-\cos x} \).
We use the quotient rule. Here, \( u = 1+\cos x \) and \( v = 1-\cos x \).
\( \frac{du}{dx} = -\sin x \) and \( \frac{dv}{dx} = \sin x \).
\( \frac{dy}{dx} = \frac{(1-\cos x)(-\sin x) - (1+\cos x)(\sin x)}{(1-\cos x)^2} \)
\( \implies \frac{dy}{dx} = \frac{-\sin x + \sin x\cos x - \sin x - \sin x\cos x}{(1-\cos x)^2} \)
\( \implies \frac{dy}{dx} = \frac{-2\sin x}{(1-\cos x)^2} \).
(iv) Let \( y = x^2 - \sqrt{1+x} \).
We can write \( \sqrt{1+x} \) as \( (1+x)^{1/2} \). So, \( y = x^2 - (1+x)^{1/2} \).
We differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}((1+x)^{1/2}) \)
\( \implies \frac{dy}{dx} = 2x - \frac{1}{2}(1+x)^{-\frac{1}{2}} \times \frac{d}{dx}(1+x) \)
\( \implies \frac{dy}{dx} = 2x - \frac{1}{2}(1+x)^{-\frac{1}{2}} \times 1 \)
\( \implies \frac{dy}{dx} = 2x - \frac{1}{2\sqrt{1+x}} \).
(v) Let \( y = \sin 2x \cos^2 x \).
We use the product rule: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \).
Here, \( u = \sin 2x \) and \( v = \cos^2 x \).
\( \frac{du}{dx} = \cos 2x \times \frac{d}{dx}(2x) = 2\cos 2x \).
\( \frac{dv}{dx} = 2\cos x \times \frac{d}{dx}(\cos x) = 2\cos x (-\sin x) = -2\sin x\cos x \).
\( \frac{dy}{dx} = (\sin 2x)(-2\sin x\cos x) + (\cos^2 x)(2\cos 2x) \)
We know that \( \sin 2x = 2\sin x\cos x \).
\( \implies \frac{dy}{dx} = (2\sin x\cos x)(-2\sin x\cos x) + 2\cos^2 x \cos 2x \)
\( \implies \frac{dy}{dx} = -4\sin^2 x\cos^2 x + 2\cos^2 x \cos 2x \)
\( \implies \frac{dy}{dx} = -\sin^2 2x + 2\cos^2 x \cos 2x \).
Using trigonometric identities for \( \sin^2 x \) and \( \cos^2 x \):
\( \sin^2 x = \frac{1-\cos 2x}{2} \) and \( \cos^2 x = \frac{1+\cos 2x}{2} \).
So, \( \sin^2 2x = \frac{1-\cos 4x}{2} \).
\( \implies \frac{dy}{dx} = -\left(\frac{1-\cos 4x}{2}\right) + 2\left(\frac{1+\cos 2x}{2}\right)\cos 2x \)
\( \implies \frac{dy}{dx} = -\frac{1}{2} + \frac{\cos 4x}{2} + (1+\cos 2x)\cos 2x \)
\( \implies \frac{dy}{dx} = -\frac{1}{2} + \frac{\cos 4x}{2} + \cos 2x + \cos^2 2x \)
Using \( \cos^2 2x = \frac{1+\cos 4x}{2} \):
\( \implies \frac{dy}{dx} = -\frac{1}{2} + \frac{\cos 4x}{2} + \cos 2x + \frac{1+\cos 4x}{2} \)
\( \implies \frac{dy}{dx} = -\frac{1}{2} + \frac{\cos 4x}{2} + \cos 2x + \frac{1}{2} + \frac{\cos 4x}{2} \)
\( \implies \frac{dy}{dx} = \cos 4x + \cos 2x \).
(vi) Let \( y = \tan^4 7x \).
We can write this as \( y = (\tan 7x)^4 \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = 4(\tan 7x)^{4-1} \times \frac{d}{dx}(\tan 7x) \)
\( \implies \frac{dy}{dx} = 4\tan^3 7x \times (\sec^2 7x) \times \frac{d}{dx}(7x) \)
\( \implies \frac{dy}{dx} = 4\tan^3 7x \sec^2 7x \times 7 \)
\( \implies \frac{dy}{dx} = 28\tan^3 7x \sec^2 7x \).
(vii) Let \( y = \frac{1}{\sin x+\cos x} \).
We can write this as \( y = (\sin x + \cos x)^{-1} \).
We differentiate both sides with respect to x using the chain rule.
\( \frac{dy}{dx} = -1(\sin x + \cos x)^{-1-1} \times \frac{d}{dx}(\sin x + \cos x) \)
\( \implies \frac{dy}{dx} = -(\sin x + \cos x)^{-2} \times (\cos x - \sin x) \)
\( \implies \frac{dy}{dx} = -\frac{(\cos x - \sin x)}{(\sin x + \cos x)^2} \)
\( \implies \frac{dy}{dx} = \frac{\sin x - \cos x}{(\sin x + \cos x)^2} \).
(viii) Let \( y = \frac{1+\cos x}{x} \).
We use the quotient rule. Here, \( u = 1+\cos x \) and \( v = x \).
\( \frac{du}{dx} = -\sin x \) and \( \frac{dv}{dx} = 1 \).
\( \frac{dy}{dx} = \frac{(x)(-\sin x) - (1+\cos x)(1)}{x^2} \)
\( \implies \frac{dy}{dx} = \frac{-x\sin x - 1 - \cos x}{x^2} \).
In simple words: For fractions, you use the quotient rule. For things multiplied together, you use the product rule. For powers of functions, you use the chain rule. Sometimes, you need to use more than one rule in a single problem, like the product rule combined with the chain rule. Simplifying the expression first can sometimes make the problem much easier.

🎯 Exam Tip: When using the quotient rule or product rule, carefully identify your 'u' and 'v' functions and their derivatives. Pay attention to signs, especially with derivatives of trigonometric functions, to avoid common errors.

Question 6. Given that \( y = \frac{\sin x-\cos x}{\sin x+\cos x} \), show that \( \frac{d y}{d x}=1+y^2 \).
Answer:
Given \( y = \frac{\sin x-\cos x}{\sin x+\cos x} \).
We need to find \( \frac{dy}{dx} \) using the quotient rule, where \( u = \sin x-\cos x \) and \( v = \sin x+\cos x \).
\( \frac{du}{dx} = \cos x-(-\sin x) = \cos x+\sin x \).
\( \frac{dv}{dx} = \cos x-\sin x \).
Using the quotient rule: \( \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \).
\( \frac{dy}{dx} = \frac{(\sin x+\cos x)(\cos x+\sin x) - (\sin x-\cos x)(\cos x-\sin x)}{(\sin x+\cos x)^2} \)
\( \implies \frac{dy}{dx} = \frac{(\sin x+\cos x)^2 - (\sin x-\cos x)^2}{(\sin x+\cos x)^2} \)
Now, we expand the numerator:
\( (\sin x+\cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x\cos x = 1 + 2\sin x\cos x \).
\( (\sin x-\cos x)^2 = \sin^2 x + \cos^2 x - 2\sin x\cos x = 1 - 2\sin x\cos x \).
Substituting these back into the expression for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{(1 + 2\sin x\cos x) - (1 - 2\sin x\cos x)}{(\sin x+\cos x)^2} \)
\( \implies \frac{dy}{dx} = \frac{1 + 2\sin x\cos x - 1 + 2\sin x\cos x}{(\sin x+\cos x)^2} \)
\( \implies \frac{dy}{dx} = \frac{4\sin x\cos x}{(\sin x+\cos x)^2} \).
This doesn't directly look like \( 1+y^2 \). Let's go back to an earlier step:
\( \frac{dy}{dx} = \frac{(\sin x+\cos x)^2 + (\sin x-\cos x)^2 - 2(\sin x-\cos x)^2}{(\sin x+\cos x)^2} \)
Instead, we can split the fraction:
\( \frac{dy}{dx} = \frac{(\sin x+\cos x)^2}{(\sin x+\cos x)^2} + \frac{(\sin x-\cos x)^2}{(\sin x+\cos x)^2} \)
This step is incorrect, as the numerator was \( A^2 - B^2 \), not \( A^2 + B^2 \). Let's restart the expansion of the numerator:
Numerator: \( (\sin x+\cos x)^2 - (\sin x-\cos x)^2 \)
Using the identity \( (A+B)^2 - (A-B)^2 = 4AB \):
Here, \( A = \sin x \) and \( B = \cos x \).
Numerator = \( 4\sin x\cos x \).
So, \( \frac{dy}{dx} = \frac{4\sin x\cos x}{(\sin x+\cos x)^2} \).
Now, we need to show that this is equal to \( 1+y^2 \).
We know \( y = \frac{\sin x-\cos x}{\sin x+\cos x} \).
\( y^2 = \left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)^2 = \frac{(\sin x-\cos x)^2}{(\sin x+\cos x)^2} \).
So, \( 1+y^2 = 1 + \frac{(\sin x-\cos x)^2}{(\sin x+\cos x)^2} \)
\( \implies 1+y^2 = \frac{(\sin x+\cos x)^2 + (\sin x-\cos x)^2}{(\sin x+\cos x)^2} \).
Expanding the numerator:
\( (\sin x+\cos x)^2 + (\sin x-\cos x)^2 = (\sin^2 x+\cos^2 x+2\sin x\cos x) + (\sin^2 x+\cos^2 x-2\sin x\cos x) \)
\( = (1+2\sin x\cos x) + (1-2\sin x\cos x) \)
\( = 1+2\sin x\cos x+1-2\sin x\cos x = 2 \).
So, \( 1+y^2 = \frac{2}{(\sin x+\cos x)^2} \).
This is not equal to \( \frac{4\sin x\cos x}{(\sin x+\cos x)^2} \). There must be a mistake in the expansion or an identity I missed. Let's re-examine the numerator from \( \frac{dy}{dx} \).
\( (\sin x+\cos x)(\cos x+\sin x) - (\sin x-\cos x)(\cos x-\sin x) \)
Numerator = \( (\sin x+\cos x)^2 - (\sin x-\cos x)(-(\sin x-\cos x)) \)
Numerator = \( (\sin x+\cos x)^2 + (\sin x-\cos x)^2 \). This is where the error was in my manual trace, the OCR was correct in its steps. My expansion of \( \frac{dv}{dx} \) as \( \cos x - \sin x \) was correct for the quotient rule but the source calculation shows:
\( \frac{dy}{dx} = \frac{(\sin x+\cos x)(\cos x+\sin x) - (\sin x-\cos x)(\cos x-\sin x)}{(\sin x+\cos x)^2} \)
\( \implies \frac{dy}{dx} = \frac{(\sin x+\cos x)^2 - [-(\sin x-\cos x)(\sin x-\cos x)]}{(\sin x+\cos x)^2} \)
\( \implies \frac{dy}{dx} = \frac{(\sin x+\cos x)^2 + (\sin x-\cos x)^2}{(\sin x+\cos x)^2} \)
Now we expand the numerator using \( (A+B)^2 + (A-B)^2 = 2(A^2+B^2) \):
Numerator = \( (\sin x+\cos x)^2 + (\sin x-\cos x)^2 = 2(\sin^2 x + \cos^2 x) = 2(1) = 2 \).
So, \( \frac{dy}{dx} = \frac{2}{(\sin x+\cos x)^2} \).
From the question, we need to show \( \frac{dy}{dx} = 1+y^2 \).
We calculate \( 1+y^2 \):
\( 1+y^2 = 1 + \left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)^2 \)
\( \implies 1+y^2 = 1 + \frac{(\sin x-\cos x)^2}{(\sin x+\cos x)^2} \)
\( \implies 1+y^2 = \frac{(\sin x+\cos x)^2 + (\sin x-\cos x)^2}{(\sin x+\cos x)^2} \)
Using the identity \( (A+B)^2 + (A-B)^2 = 2(A^2+B^2) \), the numerator becomes \( 2(\sin^2 x + \cos^2 x) = 2(1) = 2 \).
\( \implies 1+y^2 = \frac{2}{(\sin x+\cos x)^2} \).
Since both \( \frac{dy}{dx} \) and \( 1+y^2 \) simplify to the same expression, we have shown that \( \frac{dy}{dx} = 1+y^2 \).
In simple words: First, we differentiate the given function \( y \) using the division rule for derivatives. Then, we take the original function \( y \), square it, and add 1. We simplify both results. If both parts simplify to the same thing, it means the equation is proven true. It's like showing two different math puzzles have the same answer.

🎯 Exam Tip: For "show that" questions, always start with one side (usually the more complex one, like \( \frac{dy}{dx} \) in this case) and work towards the other side. Do not assume the result is true from the start. Utilize trigonometric identities like \( \sin^2 x + \cos^2 x = 1 \) and \( (A+B)^2 \pm (A-B)^2 = 2(A^2+B^2) \) or \( 4AB \) to simplify expressions efficiently.

Question 7. Differentiate with respect to x :
(i) \( (2x^2 – 1) (x^3 + 4)^3 \)
(ii) \( \frac{x^4+1}{\sqrt{1+x}} \)
(iv) \( \tan^4 2x \)
Answer:
The solutions for Question 7 are not available within the provided content range (pages 1-14) of the PDF.

🎯 Exam Tip: When differentiating products of functions, remember to use the product rule: \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \). For quotients, use the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \). Always apply the chain rule for composite functions. Breaking down complex functions into simpler parts helps manage the differentiation process.

Question 7. Differentiate with respect to x :
(i) \( (2x^2 - 1) (x^3 + 4)^3 \)
(ii) \( \frac{x^4+1}{\sqrt{1+x}} \)
(iv) \( \tan^4 2x \)

Answer:
(i) Let \( y = (2x^2 - 1) (x^3 + 4)^3 \). We use the product rule for differentiation, treating \( u = (2x^2 - 1) \) and \( v = (x^3 + 4)^3 \).
\( \frac{dy}{dx} = (2x^2 - 1) \frac{d}{dx} (x^3 + 4)^3 + (x^3 + 4)^3 \frac{d}{dx} (2x^2 - 1) \)
\( \implies \frac{dy}{dx} = (2x^2 - 1) \cdot 3(x^3 + 4)^2 \cdot (3x^2) + (x^3 + 4)^3 \cdot (4x) \)
\( \implies \frac{dy}{dx} = 9x^2(2x^2 - 1)(x^3 + 4)^2 + 4x(x^3 + 4)^3 \)
We can factor out \( x(x^3 + 4)^2 \) as a common term.
\( \implies \frac{dy}{dx} = x(x^3 + 4)^2 [9x(2x^2 - 1) + 4(x^3 + 4)] \)
\( \implies \frac{dy}{dx} = x(x^3 + 4)^2 [18x^3 - 9x + 4x^3 + 16] \)
\( \implies \frac{dy}{dx} = x(x^3 + 4)^2 [22x^3 - 9x + 16] \)
In simple words: To find the derivative of a product of two functions, we use a special rule that involves differentiating each part and adding the results. It's important to also use the chain rule for terms like \( (x^3+4)^3 \).

🎯 Exam Tip: Always identify if a function is a product or quotient before differentiating. For product rule, remember to differentiate one term at a time while keeping the other constant, then switch. Factor out common terms for a simpler final answer.

Answer:
(ii) Let \( y = \frac{x^4+1}{\sqrt{1+x}} \). We will use the quotient rule for differentiation, where \( u = x^4+1 \) and \( v = \sqrt{1+x} = (1+x)^{1/2} \).
\( \frac{dy}{dx} = \frac{\sqrt{1+x} \frac{d}{dx} (x^4+1) - (x^4+1) \frac{d}{dx} \sqrt{1+x}}{(\sqrt{1+x})^2} \)
\( \implies \frac{dy}{dx} = \frac{\sqrt{1+x} (4x^3) - (x^4+1) \cdot \frac{1}{2}(1+x)^{-1/2} \cdot 1}{1+x} \)
To simplify this expression, multiply the numerator and denominator by \( 2\sqrt{1+x} \).
\( \implies \frac{dy}{dx} = \frac{2\sqrt{1+x} \cdot \sqrt{1+x} (4x^3) - (x^4+1)}{2\sqrt{1+x}(1+x)} \)
\( \implies \frac{dy}{dx} = \frac{2(1+x)(4x^3) - (x^4+1)}{2(1+x)^{3/2}} \)
\( \implies \frac{dy}{dx} = \frac{8x^3 + 8x^4 - x^4 - 1}{2(1+x)^{3/2}} \)
\( \implies \frac{dy}{dx} = \frac{7x^4 + 8x^3 - 1}{2(1+x)^{3/2}} \)
In simple words: When a function is a fraction, we use the quotient rule to find its derivative. Remember to correctly find the derivative of both the top and bottom parts and combine them carefully.

🎯 Exam Tip: The quotient rule can be tricky, especially with radical terms. Always write \( \sqrt{f(x)} \) as \( (f(x))^{1/2} \) to easily apply the power rule and chain rule. Rationalize the numerator by multiplying by common factors to simplify the final answer.

Answer:
(ii) Let \( y = (x+1)^{-1} \). This is a simple function that can be differentiated using the chain rule with the power rule.
\( \frac{dy}{dx} = -1(x+1)^{-1-1} \cdot \frac{d}{dx}(x+1) \)
\( \implies \frac{dy}{dx} = -1(x+1)^{-2} \cdot (1) \)
\( \implies \frac{dy}{dx} = \frac{-1}{(x+1)^2} \)
This shows how quickly the derivative of a reciprocal function can be found with the chain rule.
In simple words: To differentiate this, bring the power \( -1 \) to the front, then subtract \( 1 \) from the power to get \( -2 \). Finally, multiply by the derivative of what's inside the bracket, which is just \( 1 \).

🎯 Exam Tip: For functions of the form \( \frac{1}{f(x)} \), rewriting it as \( (f(x))^{-1} \) often makes differentiation easier by applying the power rule and chain rule instead of the quotient rule. Remember that \( (f(x))^{-2} = \frac{1}{(f(x))^2} \).

Answer:
(iv) Let \( y = \tan^4 2x = (\tan 2x)^4 \). We will apply the chain rule multiple times for this differentiation.
\( \frac{dy}{dx} = 4(\tan 2x)^{4-1} \cdot \frac{d}{dx}(\tan 2x) \)
\( \implies \frac{dy}{dx} = 4\tan^3 2x \cdot (\sec^2 2x) \cdot \frac{d}{dx}(2x) \)
\( \implies \frac{dy}{dx} = 4\tan^3 2x \cdot \sec^2 2x \cdot (2) \)
\( \implies \frac{dy}{dx} = 8\tan^3 2x \sec^2 2x \)
In simple words: This uses the chain rule in layers. First, treat the whole function as 'something to the power of four'. Then, differentiate the 'tan' part. Lastly, differentiate the '2x' part.

🎯 Exam Tip: For functions like \( \sin^n(ax) \), the differentiation steps are: (1) Power rule for \( n(\sin(ax))^{n-1} \), (2) Derivative of sin is cos, so \( \cos(ax) \), (3) Derivative of ax is a. Combine these as \( na\sin^{n-1}(ax)\cos(ax) \).

Question 8. Find the gradient function \( \frac { dy }{ dx } \) for each of the following :
(i) \( y = x - 7x^2 \)
(ii) \( y = 4x^7 - 3x^3 + 5x - 11 \)

Answer:
(i) Given the function \( y = x - 7x^2 \). To find the gradient function, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(x) - \frac{d}{dx}(7x^2) \)
\( \implies \frac{dy}{dx} = 1 - 7 \cdot 2x^{2-1} \)
\( \implies \frac{dy}{dx} = 1 - 14x \)
This gives the slope of the tangent line at any point on the curve.
In simple words: We find the derivative of each term separately. The derivative of \( x \) is \( 1 \), and for \( 7x^2 \), we multiply the power by the coefficient and reduce the power by one.

🎯 Exam Tip: Remember that the derivative of \( x^n \) is \( nx^{n-1} \), and the derivative of a constant times a function is the constant times the derivative of the function.

Answer:
(ii) Given the function \( y = 4x^7 - 3x^3 + 5x - 11 \). We differentiate each term with respect to \( x \) to find the gradient function.
\( \frac{dy}{dx} = \frac{d}{dx}(4x^7) - \frac{d}{dx}(3x^3) + \frac{d}{dx}(5x) - \frac{d}{dx}(11) \)
\( \implies \frac{dy}{dx} = 4 \cdot 7x^{7-1} - 3 \cdot 3x^{3-1} + 5 \cdot 1 - 0 \)
\( \implies \frac{dy}{dx} = 28x^6 - 9x^2 + 5 \)
The derivative of a constant like \( -11 \) is always zero.
In simple words: We differentiate each part of the sum or difference individually. For each \( x \) term, we multiply the number in front by its power, then reduce the power by one. Any number by itself becomes zero.

🎯 Exam Tip: Differentiating sums and differences of functions means you differentiate each term separately. Always remember the derivative of a constant is zero, and \( \frac{d}{dx}(cx) = c \).

Question 9. Find the gradients of the following curves at the points indicated.
(i) \( y = x^2 + 5x \) at \( (0, 0) \)
(ii) \( y = (x + 1) (2x + 3) \) at \( (2, 21) \)
(iii) \( y = 2x^2 - x + \frac { 4 }{ x } \) at \( (2, 8) \)

Answer:
(i) Given the curve \( y = x^2 + 5x \). First, we find the general gradient function \( \frac{dy}{dx} \).
\( \frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(5x) \)
\( \implies \frac{dy}{dx} = 2x + 5 \)
Now, we evaluate this gradient at the point \( (0, 0) \). Substitute \( x = 0 \) into the derivative.
\( \frac{dy}{dx} \text{ at } (0, 0) = 2(0) + 5 = 5 \)
The gradient tells us the steepness of the curve at that specific point.
In simple words: First, find the derivative of the equation. This gives you a formula for the slope at any point. Then, put the \( x \)-value of the given point into this formula to find the exact slope at that spot.

🎯 Exam Tip: To find the gradient at a specific point, first differentiate the function to get the general gradient formula, and then substitute the x-coordinate of the given point into the derivative.

Answer:
(ii) Given the curve \( y = (x + 1) (2x + 3) \). First, we expand the expression to make differentiation simpler.
\( y = 2x^2 + 3x + 2x + 3 \)
\( \implies y = 2x^2 + 5x + 3 \)
Now, we find the general gradient function \( \frac{dy}{dx} \).
\( \frac{dy}{dx} = \frac{d}{dx}(2x^2) + \frac{d}{dx}(5x) + \frac{d}{dx}(3) \)
\( \implies \frac{dy}{dx} = 4x + 5 + 0 \)
\( \implies \frac{dy}{dx} = 4x + 5 \)
Next, we evaluate this gradient at the point \( (2, 21) \). Substitute \( x = 2 \) into the derivative.
\( \frac{dy}{dx} \text{ at } (2, 21) = 4(2) + 5 = 8 + 5 = 13 \)
Expanding the product before differentiating can sometimes simplify the process compared to using the product rule directly.
In simple words: First, multiply out the brackets to get a simpler equation. Then, find the derivative of this new equation. Finally, put the \( x \)-value of the given point into the derivative to get the slope at that point.

🎯 Exam Tip: For products of simple linear terms, expanding the expression first can be an easier method than using the product rule. Always simplify the function before differentiating if possible.

Answer:
(iii) Given the curve \( y = 2x^2 - x + \frac { 4 }{ x } \). We can rewrite the last term as \( 4x^{-1} \) to use the power rule.
\( y = 2x^2 - x + 4x^{-1} \)
Now, we find the general gradient function \( \frac{dy}{dx} \).
\( \frac{dy}{dx} = \frac{d}{dx}(2x^2) - \frac{d}{dx}(x) + \frac{d}{dx}(4x^{-1}) \)
\( \implies \frac{dy}{dx} = 2 \cdot 2x^{2-1} - 1 + 4 \cdot (-1)x^{-1-1} \)
\( \implies \frac{dy}{dx} = 4x - 1 - 4x^{-2} \)
\( \implies \frac{dy}{dx} = 4x - 1 - \frac{4}{x^2} \)
Next, we evaluate this gradient at the point \( (2, 8) \). Substitute \( x = 2 \) into the derivative.
\( \frac{dy}{dx} \text{ at } (2, 8) = 4(2) - 1 - \frac{4}{(2)^2} \)
\( \implies \frac{dy}{dx} = 8 - 1 - \frac{4}{4} \)
\( \implies \frac{dy}{dx} = 8 - 1 - 1 = 6 \)
This shows how important it is to express all terms in a differentiable form before starting.
In simple words: Change the fraction \( \frac{4}{x} \) to \( 4x^{-1} \). Then, find the derivative of each part. Finally, put the \( x \)-value from the point \( (2, 8) \) into the derivative to find the slope.

🎯 Exam Tip: Always rewrite terms like \( \frac{1}{x^n} \) as \( x^{-n} \) before differentiating to apply the power rule directly. Be careful with negative exponents during differentiation.

Question 10. If \( f(x) = 3x^2 - 4x \), find the value of \( a \) given that \( f'(a) = 5 \).

Answer:
Given the function \( f(x) = 3x^2 - 4x \). First, we find the derivative of \( f(x) \), denoted as \( f'(x) \).
\( f'(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(4x) \)
\( \implies f'(x) = 3 \cdot 2x^{2-1} - 4 \cdot 1 \)
\( \implies f'(x) = 6x - 4 \)
We are given that \( f'(a) = 5 \). This means we substitute \( x \) with \( a \) in the derivative.
\( f'(a) = 6a - 4 \)
Now, we set this equal to 5 and solve for \( a \).
\( 6a - 4 = 5 \)
\( \implies 6a = 5 + 4 \)
\( \implies 6a = 9 \)
\( \implies a = \frac{9}{6} \)
\( \implies a = \frac{3}{2} \)
This problem connects the derivative to finding a specific input value given the rate of change.
In simple words: First, find the derivative of \( f(x) \). Then, replace \( x \) with \( a \) in that derivative. Set this new expression equal to \( 5 \) and solve the equation to find what \( a \) is.

🎯 Exam Tip: This question requires two main steps: first, finding the derivative of the given function, and second, substituting the variable with 'a' and solving the resulting equation for 'a'. Don't forget to simplify the fraction for the final answer.

Question 11. Differentiate from first principle.
(i) \( 3x \)
(ii) \( (x + 1) (2x - 3) \)
(iii) \( \frac{2-x}{4+3x} \)
(iv) \( x^{-3/4} \)
(v) \( \sqrt{x}+\frac{1}{\sqrt{x}} (x>0) \)

Answer:
(i) Let \( f(x) = 3x \). To differentiate from the first principle, we use the limit definition: \( f'(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} \).
First, find \( f(x + \delta x) \):
\( f(x + \delta x) = 3(x + \delta x) = 3x + 3\delta x \)
Now substitute into the limit formula:
\( f'(x) = \lim_{\delta x \to 0} \frac{(3x + 3\delta x) - 3x}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{3\delta x}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} 3 \)
\( \implies f'(x) = 3 \)
Differentiating from the first principle helps us understand the fundamental definition of a derivative.
In simple words: We find the derivative using the original definition, which involves a limit. For \( 3x \), it simplifies to just \( 3 \) because the slope of a straight line is constant.

🎯 Exam Tip: When differentiating from first principles, ensure you correctly set up \( f(x + \delta x) \) and carefully simplify the expression \( \frac{f(x + \delta x) - f(x)}{\delta x} \) before evaluating the limit. Look for terms that cancel out or allow \( \delta x \) to be factored out.

Answer:
(ii) Let \( f(x) = (x + 1)(2x - 3) \). First, we expand the function:
\( f(x) = 2x^2 - 3x + 2x - 3 \)
\( \implies f(x) = 2x^2 - x - 3 \)
Next, find \( f(x + \delta x) \):
\( f(x + \delta x) = 2(x + \delta x)^2 - (x + \delta x) - 3 \)
\( \implies f(x + \delta x) = 2(x^2 + 2x\delta x + (\delta x)^2) - x - \delta x - 3 \)
\( \implies f(x + \delta x) = 2x^2 + 4x\delta x + 2(\delta x)^2 - x - \delta x - 3 \)
Now, substitute into the limit definition:
\( f'(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{(2x^2 + 4x\delta x + 2(\delta x)^2 - x - \delta x - 3) - (2x^2 - x - 3)}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{2x^2 + 4x\delta x + 2(\delta x)^2 - x - \delta x - 3 - 2x^2 + x + 3}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{4x\delta x + 2(\delta x)^2 - \delta x}{\delta x} \)
Factor out \( \delta x \) from the numerator:
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{\delta x (4x + 2\delta x - 1)}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} (4x + 2\delta x - 1) \)
Finally, apply the limit \( \delta x \to 0 \):
\( \implies f'(x) = 4x + 2(0) - 1 \)
\( \implies f'(x) = 4x - 1 \)
Expanding the product at the beginning helps manage the terms in the limit calculation.
In simple words: First, multiply the brackets to simplify \( f(x) \). Then, find \( f(x + \delta x) \). Put these into the limit formula and cancel out terms. Finally, take the limit as \( \delta x \) goes to zero to get the derivative.

🎯 Exam Tip: For polynomial functions, expanding the terms before applying the first principle definition can simplify the algebra significantly. Remember to cancel common \( \delta x \) terms before evaluating the limit.

Answer:
(iii) Let \( f(x) = \frac{2-x}{4+3x} \). Using the first principle definition, we need to find \( f(x + \delta x) \):
\( f(x + \delta x) = \frac{2 - (x + \delta x)}{4 + 3(x + \delta x)} = \frac{2 - x - \delta x}{4 + 3x + 3\delta x} \)
Now, substitute into the limit formula:
\( f'(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{\frac{2 - x - \delta x}{4 + 3x + 3\delta x} - \frac{2-x}{4+3x}}{\delta x} \)
Combine the fractions in the numerator using a common denominator:
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{1}{\delta x} \left[ \frac{(2 - x - \delta x)(4+3x) - (2-x)(4+3x+3\delta x)}{(4+3x+3\delta x)(4+3x)} \right] \)
Expand and simplify the numerator:
Numerator: \( (8 + 6x - 4x - 3x^2 - 4\delta x - 3x\delta x) - (8 + 6x + 6\delta x - 4x - 3x^2 - 3x\delta x) \)
Numerator: \( (8 + 2x - 3x^2 - 4\delta x - 3x\delta x) - (8 + 2x - 3x^2 + 6\delta x - 3x\delta x) \)
Canceling terms in the numerator: \( -4\delta x - 6\delta x = -10\delta x \)
Substitute the simplified numerator back:
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{1}{\delta x} \left[ \frac{-10\delta x}{(4+3x+3\delta x)(4+3x)} \right] \)
Cancel \( \delta x \):
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{-10}{(4+3x+3\delta x)(4+3x)} \)
Apply the limit \( \delta x \to 0 \):
\( \implies f'(x) = \frac{-10}{(4+3x+3(0))(4+3x)} \)
\( \implies f'(x) = \frac{-10}{(4+3x)^2} \)
First principle differentiation of rational functions requires careful algebraic manipulation to combine fractions before applying the limit.
In simple words: For a fraction, replace \( x \) with \( (x + \delta x) \) in the original function. Then subtract the original function, divide by \( \delta x \), and find a common denominator. Simplify the top part until \( \delta x \) can be cancelled out, then let \( \delta x \) become zero.

🎯 Exam Tip: Differentiating rational functions from first principles often involves finding a common denominator and expanding terms carefully. Pay close attention to signs during expansion and cancellation to avoid errors. The numerator should simplify to a term containing \( \delta x \).

Answer:
(iv) Let \( f(x) = x^{-3/4} \). To differentiate from first principles, we use the binomial expansion for \( (1+z)^n \approx 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots \).
\( f(x + \delta x) = (x + \delta x)^{-3/4} = x^{-3/4} \left(1 + \frac{\delta x}{x}\right)^{-3/4} \)
Using the binomial expansion \( (1+z)^n \approx 1 + nz \):
\( f(x + \delta x) \approx x^{-3/4} \left(1 + \left(-\frac{3}{4}\right)\frac{\delta x}{x} \right) \)
\( \implies f(x + \delta x) \approx x^{-3/4} - \frac{3}{4} x^{-3/4} \frac{\delta x}{x} = x^{-3/4} - \frac{3}{4} x^{-7/4} \delta x \)
Now, substitute into the limit formula:
\( f'(x) = \lim_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{(x^{-3/4} - \frac{3}{4} x^{-7/4} \delta x) - x^{-3/4}}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{- \frac{3}{4} x^{-7/4} \delta x}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \left( - \frac{3}{4} x^{-7/4} \right) \)
\( \implies f'(x) = - \frac{3}{4} x^{-7/4} \)
The binomial expansion is a useful tool when dealing with fractional powers in first principles.
In simple words: For this function with a fractional power, we use a special expansion method (binomial series) for \( f(x + \delta x) \). This helps to simplify the expression before applying the limit. After canceling \( \delta x \), the derivative is found directly.

🎯 Exam Tip: For functions with fractional or negative powers, you can use the binomial approximation \( (x+\delta x)^n = x^n(1+\frac{\delta x}{x})^n \approx x^n(1+n\frac{\delta x}{x}) \) to simplify \( f(x+\delta x) \) for first principles. This avoids complex algebraic manipulations with radicals.

Answer:
(v) Let \( f(x) = \sqrt{x} + \frac{1}{\sqrt{x}} = x^{1/2} + x^{-1/2} \). We can find the derivative of each term separately using first principles.
For \( y = \sqrt{x} \):
\( f'(x) = \lim_{\delta x \to 0} \frac{\sqrt{x+\delta x} - \sqrt{x}}{\delta x} \)
Multiply by the conjugate:
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{(\sqrt{x+\delta x} - \sqrt{x})(\sqrt{x+\delta x} + \sqrt{x})}{\delta x(\sqrt{x+\delta x} + \sqrt{x})} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{(x+\delta x) - x}{\delta x(\sqrt{x+\delta x} + \sqrt{x})} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{\delta x}{\delta x(\sqrt{x+\delta x} + \sqrt{x})} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{1}{\sqrt{x+\delta x} + \sqrt{x}} \)
\( \implies f'(x) = \frac{1}{\sqrt{x+0} + \sqrt{x}} = \frac{1}{2\sqrt{x}} \)
For \( y = \frac{1}{\sqrt{x}} = x^{-1/2} \):
\( f'(x) = \lim_{\delta x \to 0} \frac{(x+\delta x)^{-1/2} - x^{-1/2}}{\delta x} \)
Factor out \( x^{-1/2} \):
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{x^{-1/2}((1+\frac{\delta x}{x})^{-1/2} - 1)}{\delta x} \)
Using binomial expansion \( (1+z)^n \approx 1+nz \):
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{x^{-1/2}((1 - \frac{1}{2}\frac{\delta x}{x}) - 1)}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} \frac{x^{-1/2}(-\frac{1}{2}\frac{\delta x}{x})}{\delta x} \)
\( \implies f'(x) = \lim_{\delta x \to 0} -\frac{1}{2} x^{-1/2} x^{-1} \)
\( \implies f'(x) = -\frac{1}{2} x^{-3/2} \)
Combining both derivatives:
\( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2} x^{-3/2} = \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} \)
\( \implies \frac{dy}{dx} = \frac{x-1}{2x\sqrt{x}} \)
This problem demonstrates how to apply first principles to a sum of terms, often by combining the results for each term.
In simple words: This function is a sum of two parts. We find the derivative of each part using first principles separately. For \( \sqrt{x} \), we multiply by the conjugate. For \( \frac{1}{\sqrt{x}} \), we use a special expansion. Then, we add the two results to get the final answer.

🎯 Exam Tip: For sums or differences of functions, you can find the derivative of each term separately using first principles and then add or subtract the results. When dealing with square roots, multiplying by the conjugate is a common technique to simplify the numerator.