Q.Integrate the function $f'(ax+b)\,[f(ax+b)]^n$

Showing the 12 most recent of 23 on this concept.

1. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.If $\int f(x)dx = F(x) + C$, then $\dfrac{d}{dt}\displaystyle\int\limits_{g(t)}^{h(t)} f(x)dx =$ (A) $f(h(t)) - f(g(t))$ (B) $F(h(t)) - F(g(t))$ (C) $F(h(t))h'(t) - F(g(t))g'(t)$ (D) $f(h(t))h'(t) - f(g(t))g'(t)$

   ›Reveal solutionSolution

   This is the Leibniz differentiation rule for integrals with variable limits.

   Concept and Intuition

   $\int_{g(t)}^{h(t)}f(x)dx = F(h(t))-F(g(t))$ where $F'=f$. Differentiating with respect to $t$ needs the chain rule on both limits.

   Step-by-Step Solution

   1. $\displaystyle\int_{g(t)}^{h(t)} f(x)\,dx = F(h(t)) - F(g(t))$, where $F(x)=\int f(x)dx$.
   2. Differentiate w.r.t. $t$: $\dfrac{d}{dt}\big[F(h(t))-F(g(t))\big] = F'(h(t))h'(t) - F'(g(t))g'(t)$.
   3. Since $F'=f$: this equals $f(h(t))h'(t) - f(g(t))g'(t)$.

   Common Mistakes …

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2. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.Given that $\dfrac{d}{dx}\left[\int_0^{\phi(x)} f(t)\,dt\right]=\phi'(f(x))f'(x)$. If $\int_0^{x^3} f(t)\,dt = x^2\sin2\pi x$, then the value of f(8) is (A) $2\pi/3$ (B) $4\pi/3$ (C) $\pi/3$ (D) $\pi/12$

   ›Reveal solutionSolution

   Differentiate the given identity using the chain-rule form of Leibniz's theorem, then plug in $x=2$ (since $2^3=8$) to isolate $f(8)$.

   Concept and Intuition

   When the upper limit of an integral is itself a function of $x$ (here $x^3$), differentiating $\int_0^{\phi(x)} f(t)\,dt$ with respect to $x$ brings down $f(\phi(x))\cdot \phi'(x)$ by the chain rule — exactly analogous to differentiating a composite function.

   Step-by-Step Solution

   1. Given: $\displaystyle\int_0^{x^3} f(t)\,dt = x^2\sin(2\pi x)$.
   2. Differentiate both sides w.r.t. $x$. LHS: $\dfrac{d}{dx}\int_0^{x^3}f(t)dt = f(x^3)\cdot 3x^2$ (chain rule on the upper limit).
   3. RHS: $\dfrac{d}{dx}\left[x^2\sin(2\pi x)\right]=2x\sin(2\pi x)+x^2\cdot 2\pi\cos(2\pi x)$.
   4. So $3x^2 f(x^3)=2x\sin(2\pi x)+2\pi x^2\cos(2\pi x)$. …

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3. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function. If $px + my + n = 0$ is a tangent drawn to the curve $y = f(x)$ at $x = \alpha$, then at $x = 0$, $\dfrac{d}{dx}\left(f(\alpha e^{2x})\right) =$ (A) $0$ (B) $\dfrac{p}{m}$ (C) $\dfrac{-2\alpha m}{p}$ (D) $\dfrac{-2p\alpha}{m}$

   ›Reveal solutionSolution

   Read off $f'(\alpha)$ from the tangent line's slope, then apply the chain rule to $f(\alpha e^{2x})$ at $x=0$.

   Concept and Intuition

   A tangent line to $y=f(x)$ at $x=\alpha$ has slope $f'(\alpha)$. Writing the given line in slope form, $y=-\frac{p}{m}x-\frac{n}{m}$, immediately gives $f'(\alpha)=-p/m$. The rest is a direct chain-rule computation.

   Step-by-Step Solution

   1. Line $px+my+n=0 \Rightarrow y=-\frac{p}{m}x-\frac{n}{m}$, slope $=-\frac{p}{m}$.
   2. Since this line is tangent to $y=f(x)$ at $x=\alpha$: $f'(\alpha)=-\dfrac{p}{m}$.
   3. Let $h(x)=f(\alpha e^{2x})$. By the chain rule, $h'(x)=f'(\alpha e^{2x})\cdot\alpha\cdot2e^{2x}=2\alpha e^{2x}f'(\alpha e^{2x})$.
   4. At $x=0$: $h'(0)=2\alpha\cdot1\cdot f'(\alpha e^0)=2\alpha f'(\alpha)$. …

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4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $\dfrac{d}{dx}\left(A\log\left(\dfrac{\sqrt{1-x^3}+B}{\sqrt{1-x^3}+1}\right)\right) = \dfrac{1}{x\sqrt{1-x^3}}$, then $AB =$ (A) $\dfrac{1}{3}$ (B) $\dfrac{-1}{3}$ (C) $\dfrac{-2}{3}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   Matching the derivative of a log-quotient expression to $1/(x\sqrt{1-x^3})$ pins down $A=1/3$, $B=-1$, so $AB=-1/3$.

   Concept and Intuition

   This is a "guess the antiderivative form, then solve for constants" problem. Differentiate the given log expression symbolically in terms of $u=\sqrt{1-x^3}$, and choose $B$ so the resulting denominator simplifies nicely (ideally to a pure power of $x$), then fix $A$ by matching coefficients.

   Step-by-Step Solution

   1. Let $u=\sqrt{1-x^3}$, so $u' = \dfrac{-3x^2}{2\sqrt{1-x^3}} = \dfrac{-3x^2}{2u}$.
   2. $\dfrac{d}{dx}\left[A\log\left(\dfrac{u+B}{u+1}\right)\right] = A\left[\dfrac{u'}{u+B}-\dfrac{u'}{u+1}\right] = \dfrac{Au'(1-B)}{(u+B)(u+1)}$.
   3. Try $B=-1$: then $(u+B)(u+1)=(u-1)(u+1)=u^2-1 = (1-x^3)-1=-x^3$.
   4. With $B=-1$, $1-B=2$, so the expression becomes $\dfrac{2Au'}{-x^3} = \dfrac{-2A}{x^3}\cdot\dfrac{-3x^2}{2u} = \dfrac{3A}{xu}$. …

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5. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $x = \sec\theta - \cos\theta$ and $y = \sec^n\theta - \cos^n\theta$, then $(x^2 + 4)\left(\dfrac{dy}{dx}\right)^2 =$ ______ (A) $n(y+4)$ (B) $n^2(y^2+4)$ (C) $n(y+2)$ (D) $n^2(y^2+2)$

   ›Reveal solutionSolution

   This tests parametric differentiation combined with a clever algebraic identity that avoids messy trig simplification. The answer is $n^2(y^2+4)$.

   Concept and Intuition

   Both $x$ and $y$ are given as functions of the same parameter $\theta$, so we should find $dx/d\theta$ and $dy/d\theta$ separately and divide. The trick that makes this tractable is spotting that $\sec\theta\cos\theta=1$, which makes expressions like $(\sec\theta-\cos\theta)^2+4$ collapse into a perfect square $(\sec\theta+\cos\theta)^2$.

   Step-by-Step Solution

   1. $x=\sec\theta-\cos\theta$, so $x^2+4=\sec^2\theta-2+\cos^2\theta+4=\sec^2\theta+\cos^2\theta+2=(\sec\theta+\cos\theta)^2$ (using $\sec\theta\cos\theta=1$).
   2. Similarly, since $\sec^n\theta\cdot\cos^n\theta=1$, we get $y^2+4=(\sec^n\theta+\cos^n\theta)^2$.
   3. Differentiate $x$: $\dfrac{dx}{d\theta}=\sec\theta\tan\theta+\sin\theta=\tan\theta\left(\dfrac{1}{\cos\theta}\cdot\cos\theta+\dfrac{\sin\theta}{\tan\theta}\right)$; simplifying carefully gives $\dfrac{dx}{d\theta}=\tan\theta(\sec\theta+\cos\theta)$.
   4. Differentiate $y$: $\dfrac{dy}{d\theta}=n\sec^{n-1}\theta\sec\theta\tan\theta+n\cos^{n-1}\theta\sin\theta=n\tan\theta(\sec^n\theta+\cos^n\theta)$. …

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6. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $f(0) = 0$, $f'(0) = 3$, then the derivative of $y = f(f(f(f(f(x)))))$ at $x = 0$ is (A) $16$ (B) $32$ (C) $81$ (D) $243$

   ›Reveal solutionSolution

   Because $0$ is a fixed point of $f$ ($f(0)=0$), differentiating a repeated composition at $x=0$ just multiplies $f'(0)$ by itself once per composition — five times here.

   Concept and Intuition

   By the chain rule, $\dfrac{d}{dx}f(g(x))=f'(g(x))g'(x)$. For a chain of five compositions, the derivative at a point is a product of five factors of $f'$, each evaluated at the running value of the inner composition at that point. Since $f(0)=0$, the running value stays $0$ throughout, so every factor is just $f'(0)$.

   Step-by-Step Solution

   1. Let $y=f(f(f(f(f(x)))))$ (five nested $f$'s).
   2. By repeated chain rule: $y'(x)=f'\big(f(f(f(f(x))))\big)\cdot f'\big(f(f(f(x)))\big)\cdot f'\big(f(f(x))\big)\cdot f'\big(f(x)\big)\cdot f'(x)$.
   3. At $x=0$: since $f(0)=0$, we get $f(x)=0$, then $f(f(x))=f(0)=0$, and so on — every nested value at $x=0$ is $0$. …

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7. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.If $f'(x) = \sqrt{2x^2 - 1}$ and $y = f(x^3)$, then find the value of $\dfrac{dy}{dx}$ at $x = 1$. (A) -1 (B) 3 (C) 0 (D) -3

   ›Reveal solutionSolution

   Direct chain-rule application: $y=f(x^3)$ gives $y'=3x^2 f'(x^3)$, evaluated using the given formula for $f'$.

   Concept and Intuition

   When $y$ is a composition $f(g(x))$, the chain rule multiplies the outer derivative (evaluated at the inner function) by the inner function's derivative.

   Step-by-Step Solution

   1. $y=f(x^3)$, so $\dfrac{dy}{dx}=f'(x^3)\cdot\dfrac{d}{dx}(x^3)=f'(x^3)\cdot 3x^2$.
   2. At $x=1$: inner value is $x^3=1$, so we need $f'(1)=\sqrt{2(1)^2-1}=\sqrt{1}=1$.
   3. Then $\dfrac{dy}{dx}\Big|_{x=1}=1\cdot 3(1)^2=3$.

   Common Mistakes …

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8. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $f(x) = \sqrt{2^{2x}\log(3x-2)}$, then $f'(2) =$ (A) $\dfrac{4\log 2 \log 4 + 3}{\sqrt{\log 4}}$ (B) $\dfrac{8(\log 2)^2 + 3}{2\sqrt{\log 4}}$ (C) $\dfrac{8(\log 4)^2 + 3}{2\sqrt{\log 4}}$ (D) $\dfrac{8\log 2 \log 4 + 3}{2\sqrt{\log 4}}$

   ›Reveal solutionSolution

   Differentiate $f(x)=\sqrt{2^{2x}\log(3x-2)}$ using the chain rule on the square root and the product rule inside; at $x=2$ this evaluates to $\dfrac{8\log2\log4+3}{2\sqrt{\log4}}$.

   Concept and Intuition

   Whenever a function is a square root of a product, write $f=\sqrt L$ so $f' = \dfrac{L'}{2\sqrt L}$ (chain rule), then find $L'$ using the product rule since $L(x)=2^{2x}\cdot\log(3x-2)$ is a product of an exponential and a log term. This two-layer differentiation is the key technique.

   Step-by-Step Solution

   1. Let $L(x) = 2^{2x}\log(3x-2)$, so $f(x)=\sqrt{L(x)}$ and $f'(x) = \dfrac{L'(x)}{2\sqrt{L(x)}}$.
   2. Evaluate $L(2)$: $2^{2(2)}=2^4=16$; $\log(3(2)-2)=\log 4$. So $L(2) = 16\log 4$, and $\sqrt{L(2)} = \sqrt{16\log4} = 4\sqrt{\log4}$.
   3. Find $L'(x)$ by the product rule: $L'(x) = \dfrac{d}{dx}\left[2^{2x}\right]\log(3x-2) + 2^{2x}\cdot\dfrac{d}{dx}\left[\log(3x-2)\right]$.
   4. $\dfrac{d}{dx}2^{2x} = 2^{2x}\cdot\ln2\cdot 2 = 2\cdot2^{2x}\log2$ (using $\log$ as natural log consistently), i.e. $2^{2x+1}\log2$.
   5. $\dfrac{d}{dx}\log(3x-2) = \dfrac{3}{3x-2}$.
   6. So $L'(x) = 2^{2x+1}\log2\cdot\log(3x-2) + 2^{2x}\cdot\dfrac{3}{3x-2}$. …

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9. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $y = \operatorname{Sech}^{-1}\left(\dfrac{9}{9x^2 + 10}\right)$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{-18x}{\sqrt{(9x^2+10)^2 + 81}}$ (B) $\dfrac{-18x}{\sqrt{(9x^2+10)^2 - 81}}$ (C) $\dfrac{18x}{\sqrt{(9x^2+19)(9x^2+1)}}$ (D) $\dfrac{18x(9x^2+10)}{\sqrt{(9x^2+19)(9x^2+1)}}$

   ›Reveal solutionSolution

   A chain-rule differentiation of an inverse hyperbolic function, where $(9x^2+10)^2-81$ factors as a difference of squares into $(9x^2+1)(9x^2+19)$.

   Concept and Intuition

   For $y=\operatorname{sech}^{-1}u$, the standard derivative is $\dfrac{dy}{du}=\dfrac{-1}{u\sqrt{1-u^2}}$ (for $0<u<1$). The chain rule then just needs $du/dx$, and the algebra simplifies neatly because $(9x^2+10)^2-9^2$ is a difference of squares.

   Step-by-Step Solution

   1. Let $u=\dfrac{9}{9x^2+10}$. Then $\dfrac{du}{dx}=9\cdot\dfrac{-18x}{(9x^2+10)^2}=\dfrac{-162x}{(9x^2+10)^2}$.
   2. $1-u^2=1-\dfrac{81}{(9x^2+10)^2}=\dfrac{(9x^2+10)^2-81}{(9x^2+10)^2}$.
   3. Factor as a difference of squares: $(9x^2+10)^2-9^2=(9x^2+10-9)(9x^2+10+9)=(9x^2+1)(9x^2+19)$.
   4. So $\sqrt{1-u^2}=\dfrac{\sqrt{(9x^2+1)(9x^2+19)}}{9x^2+10}$.
   5. $u\sqrt{1-u^2}=\dfrac{9}{9x^2+10}\cdot\dfrac{\sqrt{(9x^2+1)(9x^2+19)}}{9x^2+10}=\dfrac{9\sqrt{(9x^2+1)(9x^2+19)}}{(9x^2+10)^2}$. …

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10. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.If $y = \sqrt{\cosh x + \sqrt{\cosh x}}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{\sinh x (2y^2 + 2\cosh x + 1)}{4y(y^2 + \cosh x)}$ (B) $\dfrac{\sinh x (2y^2 - 2\cosh x - 1)}{4y(y^2 - \cosh x)}$ (C) $\dfrac{\sinh x (1 - 2\sqrt{\cosh x})}{4y\sqrt{\cosh x}}$ (D) $\dfrac{\sinh x (1 + 2\sqrt{\cosh x})}{4y\sqrt{\cosh x}}$

    ›Reveal solutionSolution

    Square both sides to remove the outer root, differentiate implicitly, then

    tidy the resulting fraction — the answer comes out directly in terms of $y$

    and $\cosh x$, matching option (D).

    Concept and Intuition

    When $y$ is defined as a nested square root, it's usually easier to square first

    ($y^2=$ the inside) and differentiate implicitly rather than applying the chain

    rule twice directly to the nested radical — this avoids stacking two

    $\tfrac{1}{2\sqrt{\cdot}}$ factors and keeps the algebra manageable.

    Step-by-Step Solution

    1. $y=\sqrt{\cosh x+\sqrt{\cosh x}}\ \Rightarrow\ y^2=\cosh x+\sqrt{\cosh x}$.
    2. Differentiate both sides w.r.t. $x$: $2y\,\dfrac{dy}{dx}=\sinh x+\dfrac{1}{2\sqrt{\cosh x}}\cdot\sinh x=\sinh x\left(1+\dfrac{1}{2\sqrt{\cosh x}}\right)$.
    3. Combine the bracket over a common denominator: $1+\dfrac{1}{2\sqrt{\cosh x}}=\dfrac{2\sqrt{\cosh x}+1}{2\sqrt{\cosh x}}$.
    4. So $2y\,y'=\dfrac{\sinh x\,(2\sqrt{\cosh x}+1)}{2\sqrt{\cosh x}}$. …

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11. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.If $y = \mathrm{Tan}^{-1}\sqrt{x^2 - 1} + \mathrm{Sinh}^{-1}\sqrt{x^2 - 1}$, $x > 1$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{1}{x\sqrt{x^2-1}}$ (B) $\dfrac{x+1}{x\sqrt{x^2-1}}$ (C) $\dfrac{x+1}{x^2\sqrt{x^2-1}}$ (D) $\dfrac{x}{\sqrt{x^2-1}}$

    ›Reveal solutionSolution

    Both inverse-function derivatives share the same inner derivative

    $u'=x/\sqrt{x^2-1}$; the $1+u^2$ under $\mathrm{Tan}^{-1}$ and $\sqrt{1+u^2}$ under

    $\mathrm{Sinh}^{-1}$ both simplify beautifully because $u^2=x^2-1$, so $1+u^2=x^2$.

    Concept and Intuition

    Both $\mathrm{Tan}^{-1}$ and $\mathrm{Sinh}^{-1}$ have derivative formulas built around

    $1+u^2$ (as $\dfrac{1}{1+u^2}$ and $\dfrac{1}{\sqrt{1+u^2}}$ respectively). Here

    $u=\sqrt{x^2-1}$ makes $1+u^2=x^2$ exactly, a clean perfect square — this is why

    the two inverse functions are paired together in the problem, since they

    combine so tidily.

    Step-by-Step Solution

    1. Let $u=\sqrt{x^2-1}$. Then $u'=\dfrac{x}{\sqrt{x^2-1}}$ and $1+u^2 = 1+(x^2-1)=x^2$.
    2. $\dfrac{d}{dx}\mathrm{Tan}^{-1}u=\dfrac{u'}{1+u^2}=\dfrac{x/\sqrt{x^2-1}}{x^2}=\dfrac{1}{x\sqrt{x^2-1}}$.
    3. $\dfrac{d}{dx}\mathrm{Sinh}^{-1}u=\dfrac{u'}{\sqrt{1+u^2}}=\dfrac{x/\sqrt{x^2-1}}{\sqrt{x^2}}=\dfrac{x/\sqrt{x^2-1}}{x}=\dfrac{1}{\sqrt{x^2-1}}$ (since $x>1$ means $\sqrt{x^2}=x$, not $|x|$ ambiguity). …

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12. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If $y = \tanh^{-1}\sqrt{\dfrac{1-x}{1+x}}$, then $\dfrac{dy}{dx} =$ (A) $-\dfrac{1}{2\sqrt{1-x^2}}$ (B) $-\dfrac{1}{2x\sqrt{1-x^2}}$ (C) $\dfrac{2}{1+x^2}$ (D) $\dfrac{1}{2x\sqrt{1+x^2}}$

    ›Reveal solutionSolution

    Differentiating $\tanh^{-1}u$ via the chain rule and simplifying $u(1+x)=\sqrt{1-x^2}$ gives $dy/dx=-\dfrac{1}{2x\sqrt{1-x^2}}$.

    Concept and Intuition

    $\tanh^{-1}z=\dfrac12\log\dfrac{1+z}{1-z}$ has derivative $\dfrac{1}{1-z^2}$, exactly like a standard log-based inverse function. Here $z=u(x)$ is itself a composite square-root expression, so the chain rule applies twice; the algebra simplifies nicely because $u^2$ is a simple rational function of $x$.

    Step-by-Step Solution

    1. Let $u=\sqrt{\dfrac{1-x}{1+x}}$, so $y=\tanh^{-1}u$ and $\dfrac{dy}{du}=\dfrac{1}{1-u^2}$.
    2. $u^2=\dfrac{1-x}{1+x}\Rightarrow 1-u^2=1-\dfrac{1-x}{1+x}=\dfrac{(1+x)-(1-x)}{1+x}=\dfrac{2x}{1+x}$.
    3. Differentiate $u^2=\dfrac{1-x}{1+x}$ w.r.t. $x$: $2u\dfrac{du}{dx}=\dfrac{-(1+x)-(1-x)}{(1+x)^2}=\dfrac{-2}{(1+x)^2}\Rightarrow \dfrac{du}{dx}=\dfrac{-1}{u(1+x)^2}$.
    4. By the chain rule: $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}=\dfrac{1+x}{2x}\cdot\left(\dfrac{-1}{u(1+x)^2}\right)=\dfrac{-1}{2x\,u(1+x)}$. …

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