OP Malhotra Class 11 Maths Solutions Chapter 19 Differentiation Exercise 19 (D)

S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Ex 19(d)

Question 1. sin 5x
Answer: Let \( y = \sin 5x \). To find the derivative, we differentiate both sides with respect to \( x \). We use the chain rule here.
\( \frac{dy}{dx} = \frac{d}{dx} (\sin 5x) \)
\( = \cos 5x \cdot \frac{d}{dx}(5x) \)
\( = 5 \cos 5x \)
In simple words: First, find the derivative of sin (something), which is cos (something). Then, multiply this by the derivative of the "something" itself.

🎯 Exam Tip: Remember the chain rule: \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \). Apply it step-by-step for full clarity.

Question 2. cos 8x
Answer: Let \( y = \cos 8x \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} (\cos 8x) \)
\( = -\sin 8x \cdot \frac{d}{dx} (8x) \)
\( = -8 \sin 8x \)
In simple words: The derivative of cos (angle) is -sin (angle), then multiply by the derivative of the angle itself.

🎯 Exam Tip: Do not forget the negative sign when differentiating cosine functions, and always apply the chain rule for composite functions.

Question 3. sin (5x + 9)
Answer: Let \( y = \sin (5x + 9) \). To find the derivative, we differentiate both sides with respect to \( x \). We will use the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} \sin (5x + 9) \)
\( = \cos(5x + 9) \cdot \frac{d}{dx}(5x + 9) \)
\( = \cos(5x + 9) \cdot (5 \cdot 1 + 0) \)
\( = 5 \cos (5x + 9) \)
In simple words: Differentiate sin (expression) to get cos (expression), then multiply by the derivative of the expression inside the parenthesis.

🎯 Exam Tip: When taking the derivative of a sum or difference, differentiate each term separately. The derivative of a constant (like 9) is 0.

Question 4. cos (2x - 3)
Answer: Let \( y = \cos (2x - 3) \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule here.
\( \frac{dy}{dx} = \frac{d}{dx} \cos(2x - 3) \)
\( = -\sin(2x - 3) \cdot \frac{d}{dx}(2x - 3) \)
\( = -\sin(2x - 3) \cdot (2 - 0) \)
\( = -2 \sin (2x - 3) \)
In simple words: The derivative of cos (expression) is -sin (expression), multiplied by the derivative of the inner expression.

🎯 Exam Tip: Be careful with the signs; the derivative of \( \cos u \) is \( -\sin u \cdot \frac{du}{dx} \).

Question 5. tan 7x
Answer: Let \( y = \tan 7x \). To find the derivative, we differentiate both sides with respect to \( x \). We use the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} (\tan 7x) \)
\( = \sec^2 7x \cdot \frac{d}{dx} (7x) \)
\( = \sec^2 7x \cdot 7 \)
\( = 7 \sec^2 7x \)
In simple words: The derivative of tan (angle) is sec² (angle), then multiply by the derivative of the angle.

🎯 Exam Tip: Remember the standard derivative for tangent functions: \( \frac{d}{dx} (\tan u) = \sec^2 u \cdot \frac{du}{dx} \).

Question 6. cot nx
Answer: Let \( y = \cot nx \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} (\cot nx) \)
\( = -\csc^2 nx \cdot \frac{d}{dx} (nx) \)
\( = -\csc^2 nx \cdot n \)
\( = -n \csc^2 nx \)
In simple words: The derivative of cot (expression) is -cosec² (expression), multiplied by the derivative of the inner expression.

🎯 Exam Tip: Note that trigonometric functions starting with 'co' (cos, cot, csc) have negative derivatives. This helps in remembering the signs.

Question 7. tan (6x + 11)
Answer: Let \( y = \tan (6x + 11) \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} \tan (6x + 11) \)
\( = \sec^2(6x + 11) \cdot \frac{d}{dx}(6x + 11) \)
\( = \sec^2(6x + 11) \cdot (6 \cdot 1 + 0) \)
\( = 6 \sec^2 (6x + 11) \)
In simple words: Differentiate tan (expression) to get sec² (expression), then multiply by the derivative of the expression inside.

🎯 Exam Tip: Always differentiate the outer function first, keeping the inner function as is, then multiply by the derivative of the inner function.

Question 8. sin \( \frac{x}{3} \)
Answer: Let \( y = \sin \frac{x}{3} \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} \sin \frac{x}{3} \)
\( = \cos \frac{x}{3} \cdot \frac{d}{dx} \left(\frac{x}{3}\right) \)
\( = \cos \frac{x}{3} \cdot \frac{1}{3} \)
\( = \frac{1}{3} \cos \frac{x}{3} \)
In simple words: Differentiate sin (expression) to get cos (expression), and then multiply by the derivative of the expression, which is \( \frac{1}{3} \) for \( \frac{x}{3} \).

🎯 Exam Tip: \( \frac{x}{3} \) can be seen as \( \frac{1}{3}x \). Its derivative is simply \( \frac{1}{3} \).

Question 9. sec mx
Answer: Let \( y = \sec mx \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} (\sec mx) \)
\( = \sec mx \tan mx \cdot \frac{d}{dx} (mx) \)
\( = \sec mx \tan mx \cdot m \)
\( = m \sec mx \tan mx \)
In simple words: The derivative of sec (expression) is sec (expression) tan (expression), then multiply by the derivative of the inner expression.

🎯 Exam Tip: Know the derivative formulas for all six trigonometric functions and practice applying the chain rule correctly.

Question 10. sec \( \left(\frac{x}{2}-1\right) \)
Answer: Let \( y = \sec \left(\frac{x}{2}-1\right) \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} \sec\left(\frac{x}{2}-1\right) \)
\( = \sec\left(\frac{x}{2}-1\right) \tan \left(\frac{x}{2}-1\right) \cdot \frac{d}{dx} \left(\frac{x}{2}-1\right) \)
\( = \sec\left(\frac{x}{2}-1\right) \tan \left(\frac{x}{2}-1\right) \cdot \left(\frac{1}{2} - 0\right) \)
\( = \frac{1}{2} \sec \left(\frac{x}{2}-1\right) \tan \left(\frac{x}{2}-1\right) \)
In simple words: Differentiate sec (expression) to get sec (expression) tan (expression), then multiply by the derivative of the inside expression, which is \( \frac{1}{2} \).

🎯 Exam Tip: Remember that the derivative of a constant like -1 is zero, so it does not affect the derivative of the inner function.

Question 11. cosec \( \frac{2}{3}x \)
Answer: Let \( y = \csc \frac{2}{3}x \). To find the derivative, we differentiate both sides with respect to \( x \). We use the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} \left(\csc \frac{2}{3}x\right) \)
\( = -\cot \frac{2}{3}x \csc \frac{2}{3}x \cdot \frac{d}{dx} \left(\frac{2}{3}x\right) \)
\( = -\cot \frac{2}{3}x \csc \frac{2}{3}x \cdot \frac{2}{3} \)
\( = -\frac{2}{3} \cot \frac{2}{3}x \csc \frac{2x}{3} \)
In simple words: The derivative of cosec (expression) is -cot (expression) cosec (expression), multiplied by the derivative of the inner expression, which is \( \frac{2}{3} \).

🎯 Exam Tip: The derivative of \( \csc u \) is \( -\csc u \cot u \cdot \frac{du}{dx} \). Be careful with the coefficient and the negative sign.

Question 12. x sin x
Answer: Let \( y = x \sin x \). To find the derivative, we differentiate both sides with respect to \( x \). We need to apply the product rule here.
\( \frac{dy}{dx} = \frac{d}{dx} (x \sin x) \)
\( = x \frac{d}{dx} (\sin x) + \sin x \frac{d}{dx}(x) \)
\( \implies \) The product rule states: \( \frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx} \). This rule helps differentiate functions that are multiplied together.
\( = x (\cos x) + \sin x (1) \)
\( = x \cos x + \sin x \)
In simple words: When two functions are multiplied, the product rule helps you find the derivative. You take the first function multiplied by the derivative of the second, plus the second function multiplied by the derivative of the first.

🎯 Exam Tip: Identify \( u \) and \( v \) clearly before applying the product rule to avoid errors. Here, \( u=x \) and \( v=\sin x \).

Question 13. x² cos 5x
Answer: Let \( y = x^2 \cos 5x \). To find the derivative, we differentiate both sides with respect to \( x \). We use the product rule along with the chain rule.
\( \frac{dy}{dx} = x^2 \frac{d}{dx} (\cos 5x) + \cos 5x \frac{d}{dx}(x^2) \)
\( \implies \) The product rule is: \( \frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx} \). For \( \frac{d}{dx} (\cos 5x) \), we use the chain rule, getting \( -\sin 5x \cdot 5 \).
\( = x^2 (-\sin 5x) \frac{d}{dx} (5x) + \cos 5x \cdot (2x) \)
\( = x^2 (-\sin 5x) (5) + 2x \cos 5x \)
\( = -5x^2 \sin 5x + 2x \cos 5x \)
In simple words: Use the product rule for \( x^2 \) and \( \cos 5x \). Remember that differentiating \( \cos 5x \) also requires the chain rule, giving an extra factor of 5.

🎯 Exam Tip: When using the product rule, if one of the functions is a composite function, remember to apply the chain rule to it as well.

Question 14. \( \sqrt{x} \csc (5x + 7) \)
Answer: Let \( y = \sqrt{x} \csc (5x + 7) \). To find the derivative, we differentiate both sides with respect to \( x \). This requires both the product rule and the chain rule.
\( \frac{dy}{dx} = \sqrt{x} \frac{d}{dx} (\csc (5x + 7)) + \csc (5x + 7) \frac{d}{dx}(\sqrt{x}) \)
\( = \sqrt{x} \{-\cot (5x + 7) \csc (5x + 7)\} \frac{d}{dx} (5x + 7) + \csc (5x + 7) \frac{1}{2} x^{\frac{1}{2}-1} \)
\( = \sqrt{x} \{-\cot (5x + 7) \csc (5x + 7)\} (5) + \csc (5x + 7) \frac{1}{2} x^{-\frac{1}{2}} \)
\( = -5\sqrt{x} \cot (5x + 7) \csc (5x + 7) + \frac{1}{2\sqrt{x}} \csc (5x + 7) \)
In simple words: Treat \( \sqrt{x} \) as the first function and \( \csc (5x+7) \) as the second. Use the product rule. For \( \csc (5x+7) \), also use the chain rule because \( 5x+7 \) is an inner function.

🎯 Exam Tip: Express \( \sqrt{x} \) as \( x^{1/2} \) when differentiating, as it simplifies the power rule application. Be meticulous with all the negative signs and coefficients.

Question 15. \( \frac{\sin 3x}{x-6} \)
Answer: Let \( y = \frac{\sin 3x}{x-6} \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the quotient rule.
\( \frac{dy}{dx} = \frac{(x-6)\frac{d}{dx}(\sin 3x) - \sin 3x\frac{d}{dx}(x-6)}{(x-6)^2} \)
\( \implies \) The quotient rule states: \( \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}; v \neq 0 \). This rule is used when one function is divided by another.
\( = \frac{(x-6)(\cos 3x \cdot \frac{d}{dx}(3x)) - \sin 3x(1)}{(x-6)^2} \)
\( = \frac{(x-6)(\cos 3x \cdot 3) - \sin 3x}{(x-6)^2} \)
\( = \frac{3(x-6) \cos 3x - \sin 3x}{(x-6)^2} \)
In simple words: When you have one function divided by another, use the quotient rule. It means you take the bottom function times the derivative of the top, minus the top function times the derivative of the bottom, all divided by the bottom function squared.

🎯 Exam Tip: Clearly label \( u \) and \( v \) for the numerator and denominator, respectively, before applying the quotient rule to prevent common errors.

Question 16. \( \frac{\cos x}{5x} \)
Answer: Let \( y = \frac{\cos x}{5x} \). To find the derivative, we differentiate both sides with respect to \( x \). We can treat \( \frac{1}{5} \) as a constant and apply the quotient rule to \( \frac{\cos x}{x} \).
\( \frac{dy}{dx} = \frac{1}{5} \frac{d}{dx} \left(\frac{\cos x}{x}\right) \)
\( = \frac{1}{5} \left[ \frac{x \frac{d}{dx}(\cos x) - \cos x \frac{d}{dx}(x)}{x^2} \right] \)
\( = \frac{1}{5} \left[ \frac{x (-\sin x) - \cos x (1)}{x^2} \right] \)
\( = \frac{1}{5} \left[ \frac{-x \sin x - \cos x}{x^2} \right] \)
\( = \frac{-x \sin x - \cos x}{5x^2} \)
In simple words: Factor out the constant \( \frac{1}{5} \) first. Then, use the quotient rule for the remaining fraction, where the top is \( \cos x \) and the bottom is \( x \).

🎯 Exam Tip: Always factor out constants before differentiating a complex expression; it makes the process simpler and reduces calculation errors.

Question 17. \( \frac{\tan x}{2x+3} \)
Answer: Let \( y = \frac{\tan x}{2x+3} \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the quotient rule.
\( \frac{dy}{dx} = \frac{(2x+3)\frac{d}{dx}(\tan x) - \tan x\frac{d}{dx}(2x+3)}{(2x+3)^2} \)
\( \implies \) This uses the quotient rule. Remember that the derivative of \( \tan x \) is \( \sec^2 x \).
\( = \frac{(2x+3)(\sec^2 x) - \tan x(2)}{(2x+3)^2} \)
\( = \frac{(2x+3)\sec^2 x - 2 \tan x}{(2x+3)^2} \)
In simple words: Apply the quotient rule: (bottom times derivative of top) minus (top times derivative of bottom), all divided by (bottom squared).

🎯 Exam Tip: Remember that \( \frac{d}{dx}(2x+3) = 2 \). Pay close attention to parentheses when multiplying terms in the numerator.

Question 18. \( \frac{\sec (ax-b)}{x^2-2} \)
Answer: Let \( y = \frac{\sec (ax-b)}{x^2-2} \). To find the derivative, we differentiate both sides with respect to \( x \). This requires the quotient rule and the chain rule.
\( \frac{dy}{dx} = \frac{(x^2-2)\frac{d}{dx}(\sec(ax-b)) - \sec(ax-b)\frac{d}{dx}(x^2-2)}{(x^2-2)^2} \)
\( \implies \) This is the quotient rule. The derivative of \( \sec(ax-b) \) is \( \sec(ax-b)\tan(ax-b) \cdot a \) by the chain rule.
\( = \frac{(x^2-2)[\sec(ax-b)\tan(ax-b) \cdot a] - \sec(ax-b)(2x)}{(x^2-2)^2} \)
\( = \frac{a(x^2-2)\sec(ax-b)\tan(ax-b) - 2x \sec(ax-b)}{(x^2-2)^2} \)
In simple words: Use the quotient rule for the fraction. When you differentiate \( \sec(ax-b) \), you also need to use the chain rule because of \( ax-b \) inside.

🎯 Exam Tip: For complex fractions, apply the quotient rule systematically. Double-check the chain rule application for the trigonometric function in the numerator.

Question 19. sin 2x
Answer: Let \( y = \sin 2x \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} (\sin 2x) \)
\( = \cos 2x \cdot \frac{d}{dx} (2x) \)
\( = \cos 2x \cdot 2 \)
\( = 2 \cos 2x \)
In simple words: Differentiate sin (expression) to get cos (expression), then multiply by the derivative of the inner expression.

🎯 Exam Tip: Remember that the coefficient of \( x \) within the trigonometric function will be multiplied to the front when applying the chain rule.

Question 20. cos 3x
Answer: Let \( y = \cos 3x \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} (\cos 3x) \)
\( = -\sin 3x \cdot \frac{d}{dx} (3x) \)
\( = -\sin 3x \cdot 3 \)
\( = -3 \sin 3x \)
In simple words: The derivative of cos (expression) is -sin (expression), multiplied by the derivative of the inner expression.

🎯 Exam Tip: The derivative of a cosine function always includes a negative sign at the beginning. Don't forget this crucial detail.

Question 21. tan 2x
Answer: Let \( y = \tan 2x \). To find the derivative, we differentiate both sides with respect to \( x \). We use the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} (\tan 2x) \)
\( = \sec^2 2x \cdot \frac{d}{dx} (2x) \)
\( = \sec^2 2x \cdot 2 \)
\( = 2 \sec^2 2x \)
In simple words: The derivative of tan (expression) is sec² (expression), multiplied by the derivative of the inner expression.

🎯 Exam Tip: Pay attention to the power of the secant function; it's \( \sec^2 \) not just \( \sec \).

Question 22. sin \( \frac{x}{2} \)
Answer: Let \( y = \sin \frac{x}{2} \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} \sin \frac{x}{2} \)
\( = \cos \frac{x}{2} \cdot \frac{d}{dx} \left(\frac{x}{2}\right) \)
\( = \cos \frac{x}{2} \cdot \frac{1}{2} \)
\( = \frac{1}{2} \cos \frac{x}{2} \)
In simple words: Differentiate sin (expression) to get cos (expression), then multiply by the derivative of the inner expression, which is \( \frac{1}{2} \) for \( \frac{x}{2} \).

🎯 Exam Tip: Remember that \( \frac{x}{2} \) means \( \frac{1}{2}x \), so its derivative is simply \( \frac{1}{2} \).

Question 23. sec ax
Answer: Let \( y = \sec ax \). To find the derivative, we differentiate both sides with respect to \( x \). We use the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} (\sec ax) \)
\( = \sec ax \tan ax \cdot \frac{d}{dx}(ax) \)
\( = \sec ax \tan ax \cdot a \)
\( = a \sec ax \tan ax \)
In simple words: The derivative of sec (expression) is sec (expression) tan (expression), then multiply by the derivative of the inner expression.

🎯 Exam Tip: The constant multiplier 'a' from the inner function \( ax \) gets moved to the front of the entire derivative expression.

Question 24. sec (px + q)
Answer: Let \( y = \sec (px + q) \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} \sec(px + q) \)
\( = \sec(px + q) \tan (px + q) \cdot \frac{d}{dx} (px + q) \)
\( = \sec(px + q) \tan (px + q) \cdot (p \cdot 1 + 0) \)
\( = p \sec (px + q) \tan (px + q) \)
In simple words: Differentiate sec (expression) to get sec (expression) tan (expression), then multiply by the derivative of the inner expression, which is 'p'.

🎯 Exam Tip: When differentiating a linear function like \( px+q \), its derivative is simply the coefficient of \( x \), which is \( p \).

Question 25. tan (4x - 7)
Answer: Let \( y = \tan (4x - 7) \). To find the derivative, we differentiate both sides with respect to \( x \). We apply the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} \tan(4x - 7) \)
\( = \sec^2(4x - 7) \cdot \frac{d}{dx}(4x - 7) \)
\( = \sec^2(4x - 7) \cdot (4 \cdot 1 - 0) \)
\( = 4 \sec^2 (4x - 7) \)
In simple words: Differentiate tan (expression) to get sec² (expression), then multiply by the derivative of the inner expression, which is 4.

🎯 Exam Tip: Always remember that the derivative of a constant (like -7) is zero, so it does not affect the derivative of the inner function.