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

Differentiate the following functions w.r.t. x :

Question 1. \( \sin^2(x^2) \)
Answer: Let \( y = \sin^2(x^2) = (\sin x^2)^2 \)
To find the derivative, we use the chain rule.
First, differentiate the outer power function, then the sine function, and finally the inner \( x^2 \) term.
\( \frac{d y}{d x} = \frac{d}{d x}(\sin x^2)^2 \)
\( \implies \frac{d y}{d x} = 2 \sin x^2 \cdot \frac{d}{d x}(\sin x^2) \)
\( \implies \frac{d y}{d x} = 2 \sin x^2 \cdot \cos x^2 \cdot \frac{d}{d x}(x^2) \)
\( \implies \frac{d y}{d x} = 2 \sin x^2 \cos x^2 \cdot (2x) \)
We know the trigonometric identity \( 2 \sin \theta \cos \theta = \sin(2\theta) \). Here, \( \theta = x^2 \).
\( \implies \frac{d y}{d x} = (2 \sin x^2 \cos x^2) \cdot 2x \)
\( \implies \frac{d y}{d x} = \sin(2x^2) \cdot 2x \)
\( \implies \frac{d y}{d x} = 2x \sin(2x^2) \)
In simple words: To differentiate \( \sin^2(x^2) \), you first treat it as something squared, then as sine of something, and then the inner \( x^2 \). Each step uses the chain rule, and the result is simplified using a basic trigonometry identity.

🎯 Exam Tip: Remember to apply the chain rule multiple times for nested functions like this. Differentiate from the outermost function inwards, multiplying the results at each step. Also, simplify your final answer using trigonometric identities if possible.

Question 2. \( \sin \left( \frac{1+x^2}{1-x^2} \right) \)
Answer: Let \( y = \sin \left( \frac{1+x^2}{1-x^2} \right) \)
We need to differentiate both sides with respect to \( x \). We will use the chain rule, followed by the quotient rule for the fraction inside the sine function.
\( \frac{d y}{d x} = \cos \left( \frac{1+x^2}{1-x^2} \right) \cdot \frac{d}{d x} \left( \frac{1+x^2}{1-x^2} \right) \)
Now, apply the quotient rule to \( \frac{1+x^2}{1-x^2} \). The quotient rule states \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
Here, \( u = 1+x^2 \) and \( v = 1-x^2 \).
So, \( \frac{du}{dx} = 2x \) and \( \frac{dv}{dx} = -2x \).
\( \frac{d}{d x} \left( \frac{1+x^2}{1-x^2} \right) = \frac{(1-x^2)(2x) - (1+x^2)(-2x)}{(1-x^2)^2} \)
\( \implies \frac{d}{d x} \left( \frac{1+x^2}{1-x^2} \right) = \frac{2x - 2x^3 + 2x + 2x^3}{(1-x^2)^2} \)
\( \implies \frac{d}{d x} \left( \frac{1+x^2}{1-x^2} \right) = \frac{4x}{(1-x^2)^2} \)
Substitute this back into the expression for \( \frac{d y}{d x} \):
\( \frac{d y}{d x} = \cos \left( \frac{1+x^2}{1-x^2} \right) \cdot \frac{4x}{(1-x^2)^2} \)
\( \implies \frac{d y}{d x} = \frac{4x}{(1-x^2)^2} \cos \left( \frac{1+x^2}{1-x^2} \right) \)
In simple words: First, find the derivative of the outer sine function, keeping the inside part as it is. Then, find the derivative of the fraction inside the sine function using the quotient rule. Finally, multiply these two results together to get the full answer.

🎯 Exam Tip: When differentiating a composite function, always identify the "outer" and "inner" functions. Apply the chain rule by differentiating the outer function first, then multiplying by the derivative of the inner function. For fractions, the quotient rule is essential.

Question 3. \( x^2 \sin \left( \frac{1}{x} \right) \)
Answer: Let \( y = x^2 \sin \left( \frac{1}{x} \right) \)
This function is a product of two terms, \( x^2 \) and \( \sin \left( \frac{1}{x} \right) \). So, we need to use the product rule for differentiation.
The product rule states that if \( y = u \cdot v \), then \( \frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x} \).
Here, let \( u = x^2 \) and \( v = \sin \left( \frac{1}{x} \right) \).
First, find the derivatives of \( u \) and \( v \):
\( \frac{d u}{d x} = \frac{d}{d x}(x^2) = 2x \)
For \( \frac{d v}{d x} \), we use the chain rule:
\( \frac{d v}{d x} = \frac{d}{d x} \left( \sin \left( \frac{1}{x} \right) \right) = \cos \left( \frac{1}{x} \right) \cdot \frac{d}{d x} \left( \frac{1}{x} \right) \)
Since \( \frac{d}{d x} \left( \frac{1}{x} \right) = \frac{d}{d x}(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2} \).
So, \( \frac{d v}{d x} = \cos \left( \frac{1}{x} \right) \cdot \left( -\frac{1}{x^2} \right) = -\frac{1}{x^2} \cos \left( \frac{1}{x} \right) \)
Now, apply the product rule:
\( \frac{d y}{d x} = x^2 \left( -\frac{1}{x^2} \cos \left( \frac{1}{x} \right) \right) + \sin \left( \frac{1}{x} \right) (2x) \)
\( \implies \frac{d y}{d x} = -\cos \left( \frac{1}{x} \right) + 2x \sin \left( \frac{1}{x} \right) \)
In simple words: This problem asks you to find the rate of change of a multiplication of two different functions. You need to use the product rule, which involves taking the derivative of each part separately and then adding them up in a specific way. For one of the parts, you also need to use the chain rule because it's a function inside another function.

🎯 Exam Tip: When you see a multiplication of functions, the product rule is your key. Always clearly identify your 'u' and 'v' terms and their derivatives before substituting into the rule, paying close attention to any negative signs or nested functions that require the chain rule.

Question 4. \( \sqrt{a \sin^2 x + b \cos^2 x} \)
Answer: Let \( y = \sqrt{a \sin^2 x + b \cos^2 x} = (a \sin^2 x + b \cos^2 x)^{1/2} \)
We need to differentiate both sides with respect to \( x \). We will use the chain rule.
First, differentiate the outer power function \( (\cdot)^{1/2} \).
Then, differentiate the inner function \( a \sin^2 x + b \cos^2 x \).
\( \frac{d y}{d x} = \frac{1}{2} (a \sin^2 x + b \cos^2 x)^{1/2 - 1} \cdot \frac{d}{d x}(a \sin^2 x + b \cos^2 x) \)
\( \implies \frac{d y}{d x} = \frac{1}{2 \sqrt{a \sin^2 x + b \cos^2 x}} \cdot \left( \frac{d}{d x}(a \sin^2 x) + \frac{d}{d x}(b \cos^2 x) \right) \)
Now, differentiate the terms inside the parenthesis using the chain rule again:
\( \frac{d}{d x}(a \sin^2 x) = a \cdot 2 \sin x \cdot \frac{d}{d x}(\sin x) = 2a \sin x \cos x \)
\( \frac{d}{d x}(b \cos^2 x) = b \cdot 2 \cos x \cdot \frac{d}{d x}(\cos x) = 2b \cos x (-\sin x) = -2b \sin x \cos x \)
Substitute these back:
\( \frac{d y}{d x} = \frac{1}{2 \sqrt{a \sin^2 x + b \cos^2 x}} \cdot (2a \sin x \cos x - 2b \sin x \cos x) \)
Factor out \( 2 \sin x \cos x \) from the second term:
\( \frac{d y}{d x} = \frac{1}{2 \sqrt{a \sin^2 x + b \cos^2 x}} \cdot 2 \sin x \cos x (a - b) \)
Cancel the 2s:
\( \frac{d y}{d x} = \frac{(a - b) \sin x \cos x}{\sqrt{a \sin^2 x + b \cos^2 x}} \)
In simple words: To differentiate this square root function, first differentiate the square root itself, then multiply by the derivative of what's inside the square root. For the inside part, you'll need to differentiate the \( \sin^2 x \) and \( \cos^2 x \) terms, which also require the chain rule. Then, combine and simplify the terms.

🎯 Exam Tip: For square root functions, treat them as a power of \( \frac{1}{2} \) and apply the chain rule. Be careful when differentiating \( \sin^2 x \) or \( \cos^2 x \); remember it's \( (\sin x)^2 \) or \( (\cos x)^2 \), requiring the chain rule again, so \( 2 \sin x \cos x \) and \( -2 \sin x \cos x \) respectively.

Question 5. \( \cos (1 - x^2)^2 \)
Answer: Let \( y = \cos (1 - x^2)^2 \)
To differentiate this, we apply the chain rule multiple times, working from the outermost function inward.
The outermost function is cosine, then a power of 2, then \( 1 - x^2 \).
\( \frac{d y}{d x} = \frac{d}{d x} \left( \cos (1 - x^2)^2 \right) \)
Differentiate cosine first (derivative of \( \cos u \) is \( -\sin u \)):
\( \implies \frac{d y}{d x} = -\sin (1 - x^2)^2 \cdot \frac{d}{d x} \left( (1 - x^2)^2 \right) \)
Now, differentiate the term \( (1 - x^2)^2 \) using the power rule and chain rule (derivative of \( u^2 \) is \( 2u \frac{du}{dx} \)):
\( \frac{d}{d x} \left( (1 - x^2)^2 \right) = 2(1 - x^2)^{2-1} \cdot \frac{d}{d x}(1 - x^2) \)
\( \implies \frac{d}{d x} \left( (1 - x^2)^2 \right) = 2(1 - x^2) \cdot (-2x) \)
\( \implies \frac{d}{d x} \left( (1 - x^2)^2 \right) = -4x(1 - x^2) \)
Substitute this back into the expression for \( \frac{d y}{d x} \):
\( \frac{d y}{d x} = -\sin (1 - x^2)^2 \cdot (-4x(1 - x^2)) \)
\( \implies \frac{d y}{d x} = 4x(1 - x^2) \sin (1 - x^2)^2 \)
In simple words: This derivative requires you to peel back the layers of the function one by one. First, differentiate the cosine part. Then, differentiate the squared term inside. Finally, differentiate the \( 1-x^2 \) part. Multiply all these derivatives together to get the final answer.

🎯 Exam Tip: For functions with multiple nested layers, systematically apply the chain rule. Start with the outermost function, then multiply by the derivative of the next inner function, and continue this process until you reach the innermost term. Be careful with signs, especially when differentiating cosine or terms like \( (1-x^2) \).

Question 6. \( \sin^2 x \cos^3 x \)
Answer: Let \( y = \sin^2 x \cos^3 x \)
This is a product of two functions, \( \sin^2 x \) and \( \cos^3 x \). So, we apply the product rule for differentiation.
The product rule is \( \frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x} \).
Let \( u = \sin^2 x \) and \( v = \cos^3 x \).
First, find the derivatives of \( u \) and \( v \):
\( \frac{d u}{d x} = \frac{d}{d x}(\sin^2 x) = 2 \sin x \cdot \frac{d}{d x}(\sin x) = 2 \sin x \cos x \)
\( \frac{d v}{d x} = \frac{d}{d x}(\cos^3 x) = 3 \cos^2 x \cdot \frac{d}{d x}(\cos x) = 3 \cos^2 x (-\sin x) = -3 \sin x \cos^2 x \)
Now, substitute these into the product rule formula:
\( \frac{d y}{d x} = (\sin^2 x) (-3 \sin x \cos^2 x) + (\cos^3 x) (2 \sin x \cos x) \)
\( \implies \frac{d y}{d x} = -3 \sin^3 x \cos^2 x + 2 \sin x \cos^4 x \)
We can factor out common terms like \( \sin x \cos^2 x \):
\( \implies \frac{d y}{d x} = \sin x \cos^2 x (-3 \sin^2 x + 2 \cos^2 x) \)
This can also be written using \( \sin^2 x = 1 - \cos^2 x \):
\( \implies \frac{d y}{d x} = \sin x \cos^2 x (-3(1 - \cos^2 x) + 2 \cos^2 x) \)
\( \implies \frac{d y}{d x} = \sin x \cos^2 x (-3 + 3 \cos^2 x + 2 \cos^2 x) \)
\( \implies \frac{d y}{d x} = \sin x \cos^2 x (5 \cos^2 x - 3) \)
The original OCR had the form `\( \cos^2 x \sin x (2 \cos^2 x - 3 \sin^2 x) \)` which is also correct and directly from the factored step. Both forms are acceptable.
In simple words: This problem involves finding the derivative of two functions multiplied together. You need to use the product rule. Each of these functions (\( \sin^2 x \) and \( \cos^3 x \)) also requires the chain rule because they involve powers of sine and cosine. After finding all the parts, combine them and simplify the expression by factoring.

🎯 Exam Tip: When differentiating products of trigonometric powers, ensure you apply the product rule correctly. For terms like \( \sin^2 x \) or \( \cos^3 x \), remember to use the chain rule (e.g., \( \frac{d}{dx}(\sin^n x) = n \sin^{n-1} x \cos x \)) to find their individual derivatives before combining with the product rule.

Question 7. \( \cot (\sin \sqrt{x}) \)
Answer: Let \( y = \cot (\sin \sqrt{x}) \)
This function is a composition of three functions: cotangent, sine, and square root. We will use the chain rule repeatedly.
\( \frac{d y}{d x} = \frac{d}{d x} \left( \cot (\sin \sqrt{x}) \right) \)
First, differentiate cotangent (derivative of \( \cot u \) is \( -\operatorname{cosec}^2 u \)):
\( \implies \frac{d y}{d x} = -\operatorname{cosec}^2 (\sin \sqrt{x}) \cdot \frac{d}{d x} (\sin \sqrt{x}) \)
Next, differentiate sine (derivative of \( \sin v \) is \( \cos v \)):
\( \frac{d}{d x} (\sin \sqrt{x}) = \cos \sqrt{x} \cdot \frac{d}{d x} (\sqrt{x}) \)
Finally, differentiate the square root (derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \)):
\( \frac{d}{d x} (\sqrt{x}) = \frac{1}{2\sqrt{x}} \)
Substitute these derivatives back into the expression for \( \frac{d y}{d x} \):
\( \frac{d y}{d x} = -\operatorname{cosec}^2 (\sin \sqrt{x}) \cdot \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}} \)
\( \implies \frac{d y}{d x} = -\frac{\cos \sqrt{x} \operatorname{cosec}^2 (\sin \sqrt{x})}{2\sqrt{x}} \)
In simple words: To find the derivative of this function, you have to work from the outside in, taking the derivative of each layer one by one. First, differentiate the cotangent part, then the sine part, and finally the square root part. Multiply all these derivatives together to get your answer.

🎯 Exam Tip: When dealing with triple-layered composite functions, break down the chain rule into steps: outermost, then middle, then innermost. Carefully apply the correct derivative rule for each function type (cotangent, sine, square root) and multiply all results to avoid errors.

Question 8. \( \sqrt{\frac{\sec x-1}{\sec x+1}} \)
Answer: Let \( y = \sqrt{\frac{\sec x-1}{\sec x+1}} \)
Before differentiating, it's often helpful to simplify the expression using trigonometric identities.
We know that \( \sec x = \frac{1}{\cos x} \). Substitute this into the expression for \( y \):
\( y = \sqrt{\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}} \)
\( \implies y = \sqrt{\frac{\frac{1-\cos x}{\cos x}}{\frac{1+\cos x}{\cos x}}} \)
\( \implies y = \sqrt{\frac{1-\cos x}{1+\cos x}} \)
Now, use the half-angle identities: \( 1-\cos x = 2\sin^2 \frac{x}{2} \) and \( 1+\cos x = 2\cos^2 \frac{x}{2} \).
\( y = \sqrt{\frac{2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}}} \)
\( \implies y = \sqrt{\tan^2 \frac{x}{2}} \)
\( \implies y = \tan \frac{x}{2} \)
Now, differentiate \( y = \tan \frac{x}{2} \) with respect to \( x \). This requires the chain rule.
\( \frac{d y}{d x} = \frac{d}{d x} \left( \tan \frac{x}{2} \right) \)
\( \implies \frac{d y}{d x} = \sec^2 \frac{x}{2} \cdot \frac{d}{d x} \left( \frac{x}{2} \right) \)
\( \implies \frac{d y}{d x} = \sec^2 \frac{x}{2} \cdot \frac{1}{2} \)
\( \implies \frac{d y}{d x} = \frac{1}{2} \sec^2 \frac{x}{2} \)
In simple words: First, simplify the complex expression using trigonometric rules to make it much easier. Change \( \sec x \) to \( \frac{1}{\cos x} \) and use half-angle formulas to turn the whole thing into just \( \tan \frac{x}{2} \). After simplifying, take the derivative of this simpler function using the chain rule.

🎯 Exam Tip: Always look for opportunities to simplify complex trigonometric expressions using identities before differentiating. This can significantly reduce the complexity of the differentiation process. Key identities to remember include those for \( \sec x \), half-angles, and squared terms.

Question 9. \( y = \frac{\sin x+x^2}{\cot 2 x} \)
Answer: Let \( y = \frac{\sin x+x^2}{\cot 2 x} \)
We can rewrite this expression as \( y = (\sin x + x^2) \tan 2x \), since \( \frac{1}{\cot A} = \tan A \).
This is a product of two functions, \( u = (\sin x + x^2) \) and \( v = \tan 2x \). We will use the product rule for differentiation.
The product rule states that if \( y = u \cdot v \), then \( \frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x} \).
First, find the derivatives of \( u \) and \( v \):
\( \frac{d u}{d x} = \frac{d}{d x}(\sin x + x^2) = \cos x + 2x \)
For \( \frac{d v}{d x} \), we use the chain rule:
\( \frac{d v}{d x} = \frac{d}{d x}(\tan 2x) = \sec^2 2x \cdot \frac{d}{d x}(2x) \)
\( \implies \frac{d v}{d x} = \sec^2 2x \cdot 2 = 2 \sec^2 2x \)
Now, substitute these into the product rule formula:
\( \frac{d y}{d x} = (\sin x + x^2) (2 \sec^2 2x) + (\tan 2x) (\cos x + 2x) \)
\( \implies \frac{d y}{d x} = 2(\sin x + x^2) \sec^2 2x + (\cos x + 2x) \tan 2x \)
In simple words: The first step is to simplify the given fraction by changing \( \cot 2x \) in the denominator to \( \tan 2x \) in the numerator. This turns the problem into differentiating a product of two functions. Then, apply the product rule, making sure to use the chain rule when differentiating \( \tan 2x \).

🎯 Exam Tip: Always simplify trigonometric expressions before differentiating if possible, especially when fractions are involved (e.g., converting \( \frac{1}{\cot x} \) to \( \tan x \)). This can turn a complex quotient rule problem into a simpler product rule problem, reducing chances of error.

Question 10. If \( y = \frac{\sec x - \tan x}{\sec x + \tan x}, \text{ show that } \frac{d y}{d x} = -\frac{2 \sec x}{(\sec x + \tan x)^2} \)
Answer: Given \( y = \frac{\sec x - \tan x}{\sec x + \tan x} \)
We need to differentiate this function with respect to \( x \) using the quotient rule.
The quotient rule states that if \( y = \frac{u}{v} \), then \( \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^2} \).
Here, let \( u = \sec x - \tan x \) and \( v = \sec x + \tan x \).
First, find the derivatives of \( u \) and \( v \):
\( \frac{d u}{d x} = \frac{d}{d x}(\sec x - \tan x) = \sec x \tan x - \sec^2 x \)
\( \implies \frac{d u}{d x} = \sec x (\tan x - \sec x) \)
\( \frac{d v}{d x} = \frac{d}{d x}(\sec x + \tan x) = \sec x \tan x + \sec^2 x \)
\( \implies \frac{d v}{d x} = \sec x (\tan x + \sec x) \)
Now, substitute these into the quotient rule formula:
\( \frac{d y}{d x} = \frac{(\sec x + \tan x) (\sec x \tan x - \sec^2 x) - (\sec x - \tan x) (\sec x \tan x + \sec^2 x)}{(\sec x + \tan x)^2} \)
Factor out \( \sec x \) from \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) in the numerator:
\( \frac{d y}{d x} = \frac{(\sec x + \tan x) \sec x (\tan x - \sec x) - (\sec x - \tan x) \sec x (\tan x + \sec x)}{(\sec x + \tan x)^2} \)
Factor out \( \sec x \) from the entire numerator:
\( \frac{d y}{d x} = \frac{\sec x [ (\sec x + \tan x)(\tan x - \sec x) - (\sec x - \tan x)(\tan x + \sec x) ]}{(\sec x + \tan x)^2} \)
Notice that \( (\sec x + \tan x)(\tan x - \sec x) = \tan^2 x - \sec^2 x \).
And \( (\sec x - \tan x)(\tan x + \sec x) = \sec^2 x - \tan^2 x \).
Using the identity \( \sec^2 x - \tan^2 x = 1 \), we have \( \tan^2 x - \sec^2 x = -1 \).
So, the first bracket term is \( -1 \).
The second bracket term is \( 1 \).
\( \frac{d y}{d x} = \frac{\sec x [ (-1) - (1) ]}{(\sec x + \tan x)^2} \)
\( \implies \frac{d y}{d x} = \frac{\sec x (-2)}{(\sec x + \tan x)^2} \)
\( \implies \frac{d y}{d x} = -\frac{2 \sec x}{(\sec x + \tan x)^2} \)
This matches the required expression.
In simple words: To prove this, you need to use the quotient rule for differentiation, which involves taking derivatives of the top and bottom parts of the fraction. After applying the rule, simplify the expression by factoring out common terms and using the trigonometric identity \( \sec^2 x - \tan^2 x = 1 \) to get the final answer.

🎯 Exam Tip: For "show that" questions involving trigonometric functions, remember to simplify expressions carefully using identities like \( \sec^2 x - \tan^2 x = 1 \). This often makes the algebra much simpler after applying the differentiation rule (quotient rule here) and helps you reach the target expression more easily.

Question 11. If \( f(x) = 9 \sin x + \sin 3x, \text{ find } f'\left(\frac{\pi}{3}\right). \)
Answer: Given the function \( f(x) = 9 \sin x + \sin 3x \).
First, we need to find the derivative of \( f(x) \), denoted as \( f'(x) \).
Differentiate each term separately:
\( \frac{d}{d x}(9 \sin x) = 9 \cos x \)
For \( \frac{d}{d x}(\sin 3x) \), we use the chain rule (derivative of \( \sin(ax) \) is \( a \cos(ax) \)):
\( \frac{d}{d x}(\sin 3x) = \cos 3x \cdot \frac{d}{d x}(3x) = \cos 3x \cdot 3 = 3 \cos 3x \)
So, \( f'(x) = 9 \cos x + 3 \cos 3x \).
Now, we need to find the value of \( f'(x) \) when \( x = \frac{\pi}{3} \). Substitute \( x = \frac{\pi}{3} \) into \( f'(x) \):
\( f'\left(\frac{\pi}{3}\right) = 9 \cos \left(\frac{\pi}{3}\right) + 3 \cos \left(3 \cdot \frac{\pi}{3}\right) \)
\( \implies f'\left(\frac{\pi}{3}\right) = 9 \cos \left(\frac{\pi}{3}\right) + 3 \cos (\pi) \)
Recall the standard values: \( \cos \left(\frac{\pi}{3}\right) = \cos(60^\circ) = \frac{1}{2} \) and \( \cos(\pi) = \cos(180^\circ) = -1 \).
\( \implies f'\left(\frac{\pi}{3}\right) = 9 \left(\frac{1}{2}\right) + 3 (-1) \)
\( \implies f'\left(\frac{\pi}{3}\right) = \frac{9}{2} - 3 \)
To subtract, find a common denominator:
\( \implies f'\left(\frac{\pi}{3}\right) = \frac{9}{2} - \frac{6}{2} \)
\( \implies f'\left(\frac{\pi}{3}\right) = \frac{3}{2} \)
In simple words: First, find the derivative of the given function. This involves using the chain rule for the \( \sin 3x \) part. After finding the derivative, put the value of \( x = \frac{\pi}{3} \) into the derivative expression. Finally, use the known values of cosine for \( \frac{\pi}{3} \) and \( \pi \) to calculate the final numerical answer.

🎯 Exam Tip: When evaluating derivatives at a specific point, first find the general derivative function, then substitute the given value of \( x \). Remember common trigonometric values for angles like \( \frac{\pi}{3} \), \( \frac{\pi}{2} \), and \( \pi \), and always handle the chain rule correctly for terms like \( \sin(ax) \).

Question 12. Differentiate \( y = x \tan x \) and show that \( x \sin^2 x \frac{d y}{d x} = x \tan x + y \sin^2 x. \)
Answer: Given \( y = x \tan x \). Let this be equation (1).
First, differentiate \( y = x \tan x \) with respect to \( x \) using the product rule.
The product rule states that if \( y = u \cdot v \), then \( \frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x} \).
Here, let \( u = x \) and \( v = \tan x \).
\( \frac{d u}{d x} = \frac{d}{d x}(x) = 1 \)
\( \frac{d v}{d x} = \frac{d}{d x}(\tan x) = \sec^2 x \)
So, \( \frac{d y}{d x} = x (\sec^2 x) + (\tan x) (1) \)
\( \implies \frac{d y}{d x} = x \sec^2 x + \tan x \). Let this be equation (2).
Now, we need to show that \( x \sin^2 x \frac{d y}{d x} = x \tan x + y \sin^2 x \).
Let's start with the left-hand side (LHS) of the expression we need to show and substitute \( \frac{d y}{d x} \) from equation (2):
LHS \( = x \sin^2 x \frac{d y}{d x} \)
\( \implies \) LHS \( = x \sin^2 x (x \sec^2 x + \tan x) \)
Distribute \( x \sin^2 x \):
\( \implies \) LHS \( = x^2 \sin^2 x \sec^2 x + x \sin^2 x \tan x \)
We know that \( \sec^2 x = \frac{1}{\cos^2 x} \) and \( \tan x = \frac{\sin x}{\cos x} \).
\( \implies \) LHS \( = x^2 \sin^2 x \left(\frac{1}{\cos^2 x}\right) + x \sin^2 x \left(\frac{\sin x}{\cos x}\right) \)
\( \implies \) LHS \( = x^2 \left(\frac{\sin^2 x}{\cos^2 x}\right) + x \sin^2 x \tan x \)
\( \implies \) LHS \( = x^2 \tan^2 x + x \sin^2 x \tan x \)
From equation (1), we know that \( y = x \tan x \). We can substitute this back into the expression.
The term \( x^2 \tan^2 x \) can be written as \( x (x \tan x) \tan x \).
Wait, let's look at the target equation. It has \( x \tan x \) and \( y \sin^2 x \).
Let's simplify \( x^2 \tan^2 x \) and \( x \sin^2 x \tan x \) in relation to \( y \).
LHS \( = x^2 \tan^2 x + x \sin^2 x \tan x \)
From \( y = x \tan x \), we can write \( \tan x = \frac{y}{x} \).
LHS \( = x^2 \left(\frac{y}{x}\right)^2 + x \sin^2 x \left(\frac{y}{x}\right) \)
\( \implies \) LHS \( = x^2 \frac{y^2}{x^2} + y \sin^2 x \)
\( \implies \) LHS \( = y^2 + y \sin^2 x \). This is not matching the RHS in the question. Let's re-examine the OCR solution and the target proof. The OCR solution steps are: \( \implies x \sin^2 x \frac { dy }{ dx } = x \sin^2 x [ x \sec^2 x + \tan x] \) \( = x^2 \sin^2 x \sec^2 x + x \sin^2 x \tan x \) \( = x^2 \tan^2 x + \sin^2 x (x \tan x) \) \( \implies x \sin^2 x \frac { dy }{ dx } = x^2 \tan^2 x + y \sin^2 x \) (using eqn (1)) The original problem states: show that \( x \sin^2 x \frac{d y}{d x} = x \tan x + y \sin^2 x \). The OCR solution arrives at \( x^2 \tan^2 x + y \sin^2 x \). The target is \( x \tan x + y \sin^2 x \). There seems to be a mismatch. The OCR has `x² tan²x` while the target has `x tan x`. Let's re-verify the desired proof: \( x \sin^2 x \frac{d y}{d x} = x \tan x + y \sin^2 x \). We found \( \frac{d y}{d x} = x \sec^2 x + \tan x \). LHS \( = x \sin^2 x (x \sec^2 x + \tan x) \) \( = x^2 \sin^2 x \sec^2 x + x \sin^2 x \tan x \) \( = x^2 \frac{\sin^2 x}{\cos^2 x} + x \sin^2 x \tan x \) \( = x^2 \tan^2 x + x \sin^2 x \tan x \) Now, let's look at the RHS: \( x \tan x + y \sin^2 x \). Substitute \( y = x \tan x \): RHS \( = x \tan x + (x \tan x) \sin^2 x \) \( = x \tan x (1 + \sin^2 x) \) The question asks to show \( x^2 \tan^2 x + x \sin^2 x \tan x = x \tan x + x \tan x \sin^2 x \). This simplifies to \( x^2 \tan^2 x = x \tan x \). This is only true if \( x \tan x = 1 \), which is not generally true. There is an error in the question statement or the provided proof goal. The OCR solution actually derives: \( x \sin^2 x \frac{d y}{d x} = x^2 \tan^2 x + y \sin^2 x \). Given the constraint "Reproduce the worked solution's steps and method faithfully using whichever values the worked solution itself actually used to reach its final figure", and "If the source PDF/OCR contains an internal inconsistency... output ONLY the final chosen calculation as if it were always correct and intentional", I will present the proof as per the OCR's derivation which results in \( x^2 \tan^2 x + y \sin^2 x \). It seems the question text in the problem statement has a typo compared to what the solution proves. I will assume the OCR's derivation for the proof, and simplify the problem goal to match the outcome of the provided solution path. Revised interpretation of the question goal, based on the provided solution: Differentiate \( y = x \tan x \) and show that \( x \sin^2 x \frac{d y}{d x} = x^2 \tan^2 x + y \sin^2 x \). Let's proceed with this modified goal. Given \( y = x \tan x \). Let this be equation (1). First, differentiate \( y = x \tan x \) with respect to \( x \) using the product rule.
The product rule states that if \( y = u \cdot v \), then \( \frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x} \).
Here, let \( u = x \) and \( v = \tan x \).
\( \frac{d u}{d x} = \frac{d}{d x}(x) = 1 \)
\( \frac{d v}{d x} = \frac{d}{d x}(\tan x) = \sec^2 x \)
So, \( \frac{d y}{d x} = x (\sec^2 x) + (\tan x) (1) \)
\( \implies \frac{d y}{d x} = x \sec^2 x + \tan x \). Let this be equation (2).
Now, we need to show that \( x \sin^2 x \frac{d y}{d x} = x^2 \tan^2 x + y \sin^2 x \). (This is the corrected target based on the provided solution.)
Let's start with the left-hand side (LHS) of the expression:
LHS \( = x \sin^2 x \frac{d y}{d x} \)
Substitute \( \frac{d y}{d x} \) from equation (2):
\( \implies \) LHS \( = x \sin^2 x (x \sec^2 x + \tan x) \)
Distribute \( x \sin^2 x \):
\( \implies \) LHS \( = x (x \sin^2 x \sec^2 x) + x (\sin^2 x \tan x) \)
We know that \( \sec^2 x = \frac{1}{\cos^2 x} \).
\( \implies \) LHS \( = x^2 \sin^2 x \left(\frac{1}{\cos^2 x}\right) + x \sin^2 x \tan x \)
\( \implies \) LHS \( = x^2 \left(\frac{\sin^2 x}{\cos^2 x}\right) + x \sin^2 x \tan x \)
Since \( \frac{\sin x}{\cos x} = \tan x \), then \( \frac{\sin^2 x}{\cos^2 x} = \tan^2 x \).
\( \implies \) LHS \( = x^2 \tan^2 x + x \sin^2 x \tan x \)
From equation (1), \( y = x \tan x \). Substitute \( x \tan x \) with \( y \) in the second term:
\( \implies \) LHS \( = x^2 \tan^2 x + y \sin^2 x \)
This matches the right-hand side (RHS) of our corrected target equation.
Thus, \( x \sin^2 x \frac{d y}{d x} = x^2 \tan^2 x + y \sin^2 x \) is shown.
In simple words: First, you find the derivative of \( y = x \tan x \) using the product rule. Then, you take the left side of the equation you need to prove and substitute the derivative you just found. By simplifying the terms using trigonometric identities like \( \sec^2 x = \frac{1}{\cos^2 x} \) and \( \tan x = \frac{\sin x}{\cos x} \), you will reach the right side of the equation. This shows the given relationship is true.

🎯 Exam Tip: For "show that" problems, it's often best to start by calculating all necessary derivatives. Then, substitute these into one side of the equation to be proven (usually the more complex side) and simplify it step-by-step using algebraic manipulation and trigonometric identities until it matches the other side. Be aware of substituting the original function \( y \) back into the expression if it simplifies the proof.

Question 13. If \( y = a \sin x + b \cos x, \text{ show that } y^2 + \left(\frac{d y}{d x}\right)^2 = a^2+b^2 \)
Answer: Given \( y = a \sin x + b \cos x \). Let this be equation (1).
First, find the derivative of \( y \) with respect to \( x \), \( \frac{d y}{d x} \).
\( \frac{d y}{d x} = \frac{d}{d x}(a \sin x + b \cos x) \)
\( \implies \frac{d y}{d x} = a \cos x - b \sin x \). Let this be equation (2).
Now, we need to show that \( y^2 + \left(\frac{d y}{d x}\right)^2 = a^2+b^2 \).
Square both equation (1) and equation (2):
From (1): \( y^2 = (a \sin x + b \cos x)^2 \)
\( \implies y^2 = a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x \). Let this be equation (3).
From (2): \( \left(\frac{d y}{d x}\right)^2 = (a \cos x - b \sin x)^2 \)
\( \implies \left(\frac{d y}{d x}\right)^2 = a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x \). Let this be equation (4).
Now, add equation (3) and equation (4):
\( y^2 + \left(\frac{d y}{d x}\right)^2 = (a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x) + (a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x) \)
The terms \( +2ab \sin x \cos x \) and \( -2ab \sin x \cos x \) cancel each other out.
\( \implies y^2 + \left(\frac{d y}{d x}\right)^2 = a^2 \sin^2 x + b^2 \cos^2 x + a^2 \cos^2 x + b^2 \sin^2 x \)
Group the terms with \( a^2 \) and \( b^2 \):
\( \implies y^2 + \left(\frac{d y}{d x}\right)^2 = a^2 (\sin^2 x + \cos^2 x) + b^2 (\cos^2 x + \sin^2 x) \)
Using the fundamental trigonometric identity \( \sin^2 x + \cos^2 x = 1 \):
\( \implies y^2 + \left(\frac{d y}{d x}\right)^2 = a^2 (1) + b^2 (1) \)
\( \implies y^2 + \left(\frac{d y}{d x}\right)^2 = a^2 + b^2 \)
This proves the required relationship.
In simple words: First, find the derivative of the given function. Then, square both the original function \( y \) and its derivative \( \frac{dy}{dx} \). When you add these two squared results, the middle terms will cancel out, and you can use the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the expression, proving the required relationship.

🎯 Exam Tip: For problems involving trigonometric functions where you need to show a relationship between \( y \) and its derivative, often squaring both terms and adding them will lead to a simplification using identities like \( \sin^2 x + \cos^2 x = 1 \). Be careful with signs when expanding squared terms.

Question 14. If \( y = \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}}, \text{ show that } \frac{d y}{d x}+\sec^2\left(\frac{\pi}{4}-x\right) = 0. \)
Answer: Given \( y = \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} \).
First, simplify the expression for \( y \) using trigonometric identities.
We know that \( 1 = \sin^2 x + \cos^2 x \) and \( \sin 2x = 2 \sin x \cos x \).
So, \( 1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2 \).
And \( 1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x = (\cos x + \sin x)^2 \).
Substitute these into the expression for \( y \):
\( y = \sqrt{\frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2}} \)
\( \implies y = \frac{\cos x - \sin x}{\cos x + \sin x} \)
To simplify further, divide both the numerator and denominator by \( \cos x \):
\( y = \frac{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}} \)
\( \implies y = \frac{1 - \tan x}{1 + \tan x} \)
This is the tangent addition/subtraction formula for \( \tan\left(\frac{\pi}{4} - x\right) \), since \( \tan \frac{\pi}{4} = 1 \).
\( \implies y = \tan\left(\frac{\pi}{4} - x\right) \)
Now, differentiate \( y \) with respect to \( x \). Use the chain rule.
\( \frac{d y}{d x} = \frac{d}{d x} \left( \tan\left(\frac{\pi}{4} - x\right) \right) \)
The derivative of \( \tan u \) is \( \sec^2 u \frac{du}{dx} \).
\( \implies \frac{d y}{d x} = \sec^2\left(\frac{\pi}{4} - x\right) \cdot \frac{d}{d x}\left(\frac{\pi}{4} - x\right) \)
\( \implies \frac{d y}{d x} = \sec^2\left(\frac{\pi}{4} - x\right) \cdot (0 - 1) \)
\( \implies \frac{d y}{d x} = -\sec^2\left(\frac{\pi}{4} - x\right) \)
Now, we need to show that \( \frac{d y}{d x}+\sec^2\left(\frac{\pi}{4}-x\right) = 0 \).
Substitute the derivative we found:
\( -\sec^2\left(\frac{\pi}{4}-x\right) + \sec^2\left(\frac{\pi}{4}-x\right) = 0 \)
This equation is true, thus the proof is complete.
In simple words: First, simplify the given square root expression using trigonometric rules. Replace 1 with \( \sin^2 x + \cos^2 x \) and \( \sin 2x \) with \( 2 \sin x \cos x \), then simplify the fraction to \( \frac{1-\tan x}{1+\tan x} \), which is equal to \( \tan\left(\frac{\pi}{4} - x\right) \). After this, differentiate \( \tan\left(\frac{\pi}{4} - x\right) \) using the chain rule. Finally, add the derivative to \( \sec^2\left(\frac{\pi}{4}-x\right) \) to show that the sum is zero.

🎯 Exam Tip: For problems involving square roots of trigonometric expressions, look for ways to simplify the terms inside the square root using identities like \( 1 \pm \sin 2x = (\cos x \pm \sin x)^2 \) or the tangent sum/difference formulas. This often turns a complex differentiation problem into a much simpler one.

Question 15. If \( y = 2 \tan \frac{x}{2}, \text{ prove that } \frac{d y}{d x} = \frac{2}{1+\cos x}. \)
Answer: Given \( y = 2 \tan \frac{x}{2} \).
First, find the derivative of \( y \) with respect to \( x \). Use the chain rule.
\( \frac{d y}{d x} = \frac{d}{d x} \left( 2 \tan \frac{x}{2} \right) \)
\( \implies \frac{d y}{d x} = 2 \cdot \sec^2 \frac{x}{2} \cdot \frac{d}{d x} \left(\frac{x}{2}\right) \)
\( \implies \frac{d y}{d x} = 2 \cdot \sec^2 \frac{x}{2} \cdot \left(\frac{1}{2}\right) \)
\( \implies \frac{d y}{d x} = \sec^2 \frac{x}{2} \)
Now, we need to show that this derivative is equal to \( \frac{2}{1+\cos x} \).
We know the trigonometric identity \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \). So, for \( \theta = \frac{x}{2} \):
\( \frac{d y}{d x} = \frac{1}{\cos^2 \frac{x}{2}} \)
Also, we know the half-angle identity for cosine: \( \cos^2 \frac{x}{2} = \frac{1+\cos x}{2} \).
Substitute this identity into the expression for \( \frac{d y}{d x} \):
\( \frac{d y}{d x} = \frac{1}{\frac{1+\cos x}{2}} \)
\( \implies \frac{d y}{d x} = \frac{2}{1+\cos x} \)
This proves the required relationship.
In simple words: To prove this, first find the derivative of \( 2 \tan \frac{x}{2} \) using the chain rule. You will get \( \sec^2 \frac{x}{2} \). Then, use trigonometric identities: first, change \( \sec^2 \) to \( \frac{1}{\cos^2} \). Next, replace \( \cos^2 \frac{x}{2} \) with its half-angle formula \( \frac{1+\cos x}{2} \). After simplifying, you will get the target expression \( \frac{2}{1+\cos x} \).

🎯 Exam Tip: When proving trigonometric derivative identities, it's often necessary to convert the initial derivative into a form that uses the identities of the target expression. Remember key identities like \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \) and especially the half-angle formulas like \( \cos^2 \frac{x}{2} = \frac{1+\cos x}{2} \).