CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 02

Question 1. \(f(x) = \frac{2x^2 + 1}{x - 3}\) Using first principle method.
Answer: Using the first principle of derivatives, the derivative is given by:
\(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)
Substitute \(f(x + h) = \frac{2(x + h)^2 + 1}{x + h - 3}\) and \(f(x) = \frac{2x^2 + 1}{x - 3}\):
\(f'(x) = \lim_{h \to 0} \frac{1}{h} \left[ \frac{2(x + h)^2 + 1}{x + h - 3} - \frac{2x^2 + 1}{x - 3} \right]\)
Taking the common denominator inside the bracket:
\(f'(x) = \lim_{h \to 0} \frac{1}{h} \left[ \frac{(2(x^2 + 2xh + h^2) + 1)(x - 3) - (2x^2 + 1)(x + h - 3)}{(x + h - 3)(x - 3)} \right]\)
Expanding the numerator terms:
\(f'(x) = \lim_{h \to 0} \frac{2x^3 - 6x^2 + 2h^2x - 6h^2 + 2hx^2 - 6hx + x - 3 - 2x^3 - 2x^2h + 6x^2 - x + h - 3}{h(x + h - 3)(x - 3)}\)
Simplifying the numerator by cancelling out terms:
\(f'(x) = \lim_{h \to 0} \frac{2h^2x - 6h^2 - 6hx - h}{h(x + h - 3)(x - 3)}\)
Factoring out \(h\) from the numerator:
\(f'(x) = \lim_{h \to 0} \frac{h(2hx - 6h - 6x - 1)}{h(x + h - 3)(x - 3)}\)
Cancelling \(h\) and taking the limit as \(h \to 0\):
\(f'(x) = \frac{-6x - 1}{(x - 3)(x - 3)} = \frac{-6x - 1}{(x - 3)^2}\)
In simple words: We applied the definition of derivative using limits and simplified the complex algebraic fraction to get the final result.

Exam Tip: Be very careful with negative signs and algebraic expansions when subtracting rational expressions.

Question 2. \(f(x) = \sqrt{2x + 3}\) Using first principle method.
Answer: Using the first principle of derivatives:
\(f'(x) = \lim_{h \to 0} \frac{\sqrt{2(x + h) + 3} - \sqrt{2x + 3}}{h}\)
Rationalize the numerator by multiplying and dividing by the conjugate \(\sqrt{2x + 2h + 3} + \sqrt{2x + 3}\):
\(f'(x) = \lim_{h \to 0} \frac{(\sqrt{2x + 2h + 3} - \sqrt{2x + 3})(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})}{h(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})}\)
Using the identity \((a - b)(a + b) = a^2 - b^2\):
\(f'(x) = \lim_{h \to 0} \frac{(2x + 2h + 3) - (2x + 3)}{h(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})}\)
Simplifying the numerator:
\(f'(x) = \lim_{h \to 0} \frac{2h}{h(\sqrt{2x + 2h + 3} + \sqrt{2x + 3})}\)
Cancelling \(h\):
\(f'(x) = \lim_{h \to 0} \frac{2}{\sqrt{2x + 2h + 3} + \sqrt{2x + 3}}\)
Applying the limit \(h \to 0\):
\(f'(x) = \frac{2}{\sqrt{2x + 3} + \sqrt{2x + 3}} = \frac{2}{2\sqrt{2x + 3}} = \frac{1}{\sqrt{2x + 3}}\)
In simple words: Rationalization helps eliminate the square roots from the numerator so we can safely substitute \(h = 0\).

Exam Tip: Whenever square roots appear in first principle problems, always rationalize the numerator first.

Question 3. \(f(x) = x^2 \sin x\) Using first principle method.
Answer: Using the first principle of derivatives:
\(f'(x) = \lim_{h \to 0} \frac{(x + h)^2 \sin(x + h) - x^2 \sin x}{h}\)
Expanding \((x + h)^2 = x^2 + h^2 + 2xh\):
\(f'(x) = \lim_{h \to 0} \frac{(x^2 + h^2 + 2xh)\sin(x + h) - x^2 \sin x}{h}\)
Rearranging and grouping terms containing \(x^2\):
\(f'(x) = \lim_{h \to 0} \left[ \frac{x^2(\sin(x + h) - \sin x) + (h^2 + 2xh)\sin(x + h)}{h} \right]\)
Using the trigonometric identity for the difference of sines, \(\sin(A + B) - \sin B\):
\(f'(x) = \lim_{h \to 0} \left[ x^2 \frac{2 \cos\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right)}{h} + \frac{h(h + 2x)\sin(x + h)}{h} \right]\)
Adjusting the denominator for the standard limit \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(f'(x) = \lim_{h \to 0} \left[ x^2 \cos\left(\frac{2x + h}{2}\right) \frac{\sin(h/2)}{h/2} + (h + 2x)\sin(x + h) \right]\)
Applying the limit as \(h \to 0\):
\(f'(x) = x^2 \cos(x) \cdot 1 + (0 + 2x)\sin(x)\)
\(f'(x) = x^2 \cos x + 2x \sin x\)
In simple words: We separated the expression into parts, used trigonometric transformation formulas, and applied standard limit rules.

Exam Tip: Memorize the sum-to-product trigonometric formulas as they are essential for solving trigonometric limits by first principle.

Question 4. Differentiate w.r.t \(x\) (product rule): \(f(x) = (ax + b)^m (cx + d)^n\)
Answer: Using the product rule \(\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}\):
\(f'(x) = (ax + b)^m \cdot \frac{d}{dx}((cx + d)^n) + (cx + d)^n \cdot \frac{d}{dx}((ax + b)^m)\)
Applying the chain rule for powers:
\(f'(x) = (ax + b)^m \cdot n(cx + d)^{n-1} \cdot c + (cx + d)^n \cdot m(ax + b)^{m-1} \cdot a\)
Factoring out common terms \((ax + b)^{m-1}\) and \((cx + d)^{n-1}\):
\(f'(x) = (ax + b)^{m-1}(cx + d)^{n-1} [nc(ax + b) + ma(cx + d)]\)
Expanding the bracket inside:
\(f'(x) = (ax + b)^{m-1}(cx + d)^{n-1} [nacx + nbc + macx + mad]\)
Grouping terms:
\(f'(x) = (ax + b)^{m-1}(cx + d)^{n-1} [ax(nc + ma) + (nbc + mad)]\)
In simple words: We applied the product rule combined with the chain rule, then simplified the expression by taking common factors outside.

Exam Tip: Always look for common algebraic factors after differentiation to simplify your final answer.

Question 5. Differentiate w.r.t \(x\) (product rule): \(f(x) = (x + \sec x)(x - \tan x)\)
Answer: Applying the product rule \(\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}\):
\(f'(x) = (x + \sec x) \cdot \frac{d}{dx}(x - \tan x) + (x - \tan x) \cdot \frac{d}{dx}(x + \sec x)\)
Differentiating the individual standard functions (\(\frac{d}{dx}(x) = 1\), \(\frac{d}{dx}(\tan x) = \sec^2 x\), \(\frac{d}{dx}(\sec x) = \sec x \tan x\)):
\(f'(x) = (x + \sec x)(1 - \sec^2 x) + (x - \tan x)(1 + \sec x \tan x)\)
In simple words: Differentiate the first term while keeping the second fixed, and vice versa using the product rule.

Exam Tip: Memorize standard derivatives of trigonometric functions like secant and tangent to avoid mistakes during exams.

Question 6. Differentiate w.r.t \(x\) (product rule): \(f(x) = (x \sin x + \cos x) + (x \cos x - \sin x)\)
Answer: First, note that the expression is a sum of two functions, but let us treat the terms carefully or assume the product form as indicated in the standard problem format: \(f(x) = (x \sin x + \cos x)(x \cos x - \sin x)\).
Applying the product rule:
\(f'(x) = (x \sin x + \cos x) \cdot \frac{d}{dx}(x \cos x - \sin x) + (x \cos x - \sin x) \cdot \frac{d}{dx}(x \sin x + \cos x)\)
Differentiating each part using the product rule for internal terms:
\(f'(x) = (x \sin x + \cos x)[x(-\sin x) + \cos x \cdot 1 - \cos x] + (x \cos x - \sin x)[x(\cos x) + \sin x \cdot 1 - \sin x]\)
Simplifying inside brackets:
\(f'(x) = (x \sin x + \cos x)(-x \sin x) + (x \cos x - \sin x)(x \cos x)\)
Expanding the products:
\(f'(x) = -x^2 \sin^2 x - x \sin x \cos x + x^2 \cos^2 x - x \sin x \cos x\)
Grouping terms:
\(f'(x) = x^2(\cos^2 x - \sin^2 x) - 2x \sin x \cos x\)
Using double angle formulas (\(\cos^2 x - \sin^2 x = \cos(2x)\) and \(2 \sin x \cos x = \sin(2x)\)):
\(f'(x) = x^2 \cos(2x) - x \sin(2x)\)
In simple words: Apply the product rule, expand all terms, and use double-angle trigonometric identities to simplify the final result.

Exam Tip: Recognizing double angle identities (\(\cos 2x\) and \(\sin 2x\)) at the end of trigonometric derivations is crucial for getting full marks.

Question 7. Differentiate w.r.t \(x\): \(f(x) = x^{-4}(3 - 4x^{-5})\)
Answer: Applying the product rule:
\(f'(x) = x^{-4} \cdot \frac{d}{dx}(3 - 4x^{-5}) + (3 - 4x^{-5}) \cdot \frac{d}{dx}(x^{-4})\)
Differentiating using the power rule \(\frac{d}{dx}(x^n) = n x^{n-1}\):
\(f'(x) = x^{-4}(0 - 4(-5)x^{-6}) + (3 - 4x^{-5})(-4x^{-5})\)
Simplifying the terms:
\(f'(x) = x^{-4}(20x^{-6}) - (3 - 4x^{-5})(4x^{-5})\)
Multiplying out:
\(f'(x) = 20x^{-10} - 12x^{-5} + 16x^{-10}\)
Combining like powers of \(x\):
\(f'(x) = 36x^{-10} - 10x^{-5}\)
In simple words: Use the power rule for negative exponents carefully and combine like terms at the end.

Exam Tip: Be careful with signs when differentiating terms with negative powers, especially during product rule expansions.

Question 8. Differentiate w.r.t \(x\): \(f(x) = \frac{1 - \tan x}{1 + \tan x}\)
Answer: Converting tangent to sine and cosine:
\(f(x) = \frac{1 - \frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}} = \frac{\cos x - \sin x}{\cos x + \sin x}\)
Applying the quotient rule \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\):
\(f'(x) = \frac{(\cos x + \sin x) \cdot \frac{d}{dx}(\cos x - \sin x) - (\cos x - \sin x) \cdot \frac{d}{dx}(\cos x + \sin x)}{(\cos x + \sin x)^2}\)
Calculating the derivatives of the numerator components:
\(f'(x) = \frac{(\cos x + \sin x)(-\sin x - \cos x) - (\cos x - \sin x)(-\sin x + \cos x)}{(\cos x + \sin x)^2}\)
Simplifying the numerators:
\(f'(x) = \frac{-(\cos x + \sin x)^2 - (\cos x - \sin x)^2}{(\cos x + \sin x)^2}\)
Expanding squares and simplifying:
\(f'(x) = \frac{-(\cos^2 x + \sin^2 x + 2\sin x \cos x) - (\cos^2 x + \sin^2 x - 2\sin x \cos x)}{(\cos x + \sin x)^2}\)
Using \(\cos^2 x + \sin^2 x = 1\):
\(f'(x) = \frac{-(1 + \sin(2x)) - (1 - \sin(2x))}{(\cos x + \sin x)^2}\)
\(f'(x) = \frac{-2}{(\cos x + \sin x)^2}\)
In simple words: Convert to sine and cosine, apply the quotient rule, and simplify using standard trigonometric identities.

Question 9. Differentiate w.r.t \(x\): \(f(x) = \frac{x}{\sin^n x}\)
Answer: Applying the quotient rule \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\):
\(f'(x) = \frac{\sin^n x \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(\sin^n x)}{(\sin^n x)^2}\)
Differentiating each part:
\(f'(x) = \frac{\sin^n x(1) - x \cdot n \sin^{n-1} x \cdot \cos x}{\sin^{2n} x}\)
Factoring out \(\sin^{n-1} x\) from the numerator:
\(f'(x) = \frac{\sin^{n-1} x (\sin x - nx \cos x)}{\sin^{2n} x}\)
Cancelling common terms in numerator and denominator:
\(f'(x) = \frac{\sin x - nx \cos x}{\sin^{2n - n + 1} x} = \frac{\sin x - nx \cos x}{\sin^{n+1} x}\)
In simple words: Apply the quotient rule, use the chain rule for the power of sine, and cancel common powers to simplify.

Exam Tip: Look out for common exponent factors in the numerator and denominator to simplify rational trigonometric derivatives.

Question 10. \(f(x) = \frac{x^2 \cos\left(\frac{\pi}{4}\right)}{\sin x}\)
Answer: Since \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\), the function simplifies to:
\(f(x) = \frac{1}{\sqrt{2}} \cdot \frac{x^2}{\sin x}\)
Applying the quotient rule:
\(f'(x) = \frac{1}{\sqrt{2}} \left[ \frac{\sin x \cdot \frac{d}{dx}(x^2) - x^2 \cdot \frac{d}{dx}(\sin x)}{(\sin x)^2} \right]\)
Calculating the derivatives:
\(f'(x) = \frac{1}{\sqrt{2}} \left[ \frac{\sin x(2x) - x^2 \cos x}{\sin^2 x} \right]\)
\(f'(x) = \frac{1}{\sqrt{2}} \left[ \frac{2x \sin x - x^2 \cos x}{\sin^2 x} \right]\)
In simple words: Replace the constant trigonometric value first, then apply the quotient rule directly.

Exam Tip: Always evaluate constant trigonometric terms like \(\cos\left(\frac{\pi}{4}\right)\) before differentiating to make calculations simpler.