Q.If $u, v$ and $w$ are functions of $x$, then show that $\frac{d}{dx} (u \cdot v \cdot w) = \frac{du}{dx} \cdot v \cdot w + u \cdot \frac{dv}{dx} \cdot w + u \cdot v \cdot \frac{dw}{dx}$ in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Showing the 12 most recent of 23 on this concept.

1. 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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2. 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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3. 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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4. 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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5. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.If $x=3\left[\sin t-\log\left(\cot\dfrac{t}{2}\right)\right]$ and $y=6\left[\cos t+\log\left(\tan\dfrac{t}{2}\right)\right]$ then $\dfrac{dy}{dx}=$ (A) $\dfrac{2\sin^2 t}{1+\sin t \cos t}$ (B) $\dfrac{2\cos^2 t}{1+\sin 2t}$ (C) $\dfrac{2\cos^2 t}{1+\sin t\cos t}$ (D) $\dfrac{1+\cos 2t}{1+\sin 2t}$

   ›Reveal solutionSolution

   A parametric-differentiation problem built around the classic identity $\dfrac{d}{dt}\log\tan(t/2)=\csc t$; the final simplified slope is $\dfrac{2\cos^2t}{1+\sin t\cos t}$.

   Concept and Intuition

   When $x$ and $y$ are given as functions of a parameter $t$, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$. The log-tangent-half-angle term is a recurring building block whose derivative simplifies neatly to $\csc t$, which is worth memorizing to avoid a messy chain-rule expansion each time.

   Step-by-Step Solution

   1. Recall $\dfrac{d}{dt}\log\tan(t/2)=\dfrac{\sec^2(t/2)}{2\tan(t/2)}=\dfrac{1}{2\sin(t/2)\cos(t/2)}=\dfrac1{\sin t}=\csc t$. Hence $\dfrac{d}{dt}\log\cot(t/2)=-\csc t$.
   2. $x=3[\sin t-\log\cot(t/2)]\Rightarrow \dfrac{dx}{dt}=3[\cos t-(-\csc t)]=3(\cos t+\csc t)$.
   3. $y=6[\cos t+\log\tan(t/2)]\Rightarrow \dfrac{dy}{dt}=6[-\sin t+\csc t]$.
   4. $\dfrac{dy}{dx}=\dfrac{6(\csc t-\sin t)}{3(\cos t+\csc t)}=\dfrac{2(\csc t-\sin t)}{\cos t+\csc t}$. …

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6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $x = \mathrm{Sinh}^{-1}t + \log(t^2 + 1)$ and $y = \mathrm{Tan}^{-1}t + \log|t|$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{t^2 + t + 1}{2t + \sqrt{t^4 + t^2}}$ (B) $\dfrac{t^2 + t + 1}{2t + \sqrt{t^2 + 1}}$ (C) $\dfrac{t^2 + t + 1}{2t^2 + \sqrt{t^4 + t^2}}$ (D) $\dfrac{t^2 + t + 1}{2 + \sqrt{1 + t^2}}$

   ›Reveal solutionSolution

   This tests parametric differentiation $\left(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\right)$ with inverse-hyperbolic and inverse-trig terms; the answer is option (C).

   Concept and Intuition

   When both $x$ and $y$ are given in terms of a parameter $t$, we find $dy/dx$ as the ratio of the two derivatives with respect to $t$, using $\frac{d}{dt}\mathrm{Sinh}^{-1}t=\frac{1}{\sqrt{1+t^2}}$ and $\frac{d}{dt}\mathrm{Tan}^{-1}t=\frac{1}{1+t^2}$.

   Step-by-Step Solution

   1. $\dfrac{dx}{dt} = \dfrac{1}{\sqrt{1+t^2}} + \dfrac{2t}{t^2+1}$ (since $\frac{d}{dt}\log(t^2+1)=\frac{2t}{t^2+1}$).
   2. $\dfrac{dy}{dt} = \dfrac{1}{1+t^2} + \dfrac1t = \dfrac{t + (1+t^2)}{t(1+t^2)} = \dfrac{t^2+t+1}{t(1+t^2)}$.
   3. Write $\dfrac{dx}{dt}$ over common denominator $1+t^2$: $\dfrac{dx}{dt} = \dfrac{\sqrt{1+t^2}+2t}{1+t^2}$. …

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7. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $u = \sin\left(\dfrac{x}{y}\right)$, $x = e^t$ and $y = t^2$, then $t^6 \left(\dfrac{du}{dt}\right)^2 \div e^{2t}(t-2)^2 =$ (A) $2u$ (B) $u^2$ (C) $1 - u^2$ (D) $\cos u$

   ›Reveal solutionSolution

   Differentiating $u=\sin(x/y)$ with $x=e^t,y=t^2$ and simplifying the given combination leaves exactly $\cos^2(x/y)=1-u^2$.

   Concept and Intuition

   Compute $du/dt$ via the chain rule, then see how the algebraic combination in the question is designed to cancel everything except $\cos^2(x/y)$.

   Step-by-Step Solution

   1. $\dfrac{x}{y}=\dfrac{e^t}{t^2}$. $\dfrac{d}{dt}\left(\dfrac{e^t}{t^2}\right) = \dfrac{e^tt^2-e^t\cdot2t}{t^4}=\dfrac{e^t(t-2)}{t^3}$.
   2. $\dfrac{du}{dt}=\cos\left(\dfrac{x}{y}\right)\cdot\dfrac{e^t(t-2)}{t^3}$.
   3. $\left(\dfrac{du}{dt}\right)^2 = \cos^2\left(\dfrac{x}{y}\right)\cdot\dfrac{e^{2t}(t-2)^2}{t^6}$. …

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8. 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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9. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $x = \sqrt{2^{\operatorname{cosec}^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$, $|t| \geq 1$ then $\dfrac{dy}{dx} =$ (A) $\dfrac{x}{y}$ (B) $\dfrac{y}{x}$ (C) $-\dfrac{y}{x}$ (D) $-\dfrac{x}{y}$

   ›Reveal solutionSolution

   The identity $\operatorname{cosec}^{-1}t+\sec^{-1}t=\pi/2$ lets both $x$ and $y$ be written as exponentials of a single parameter $u$, so $dy/dx$ follows from parametric differentiation. Answer: $-\dfrac yx$.

   Concept and Intuition

   When $x$ and $y$ are both given as functions of a common (possibly hidden) parameter — here through the complementary inverse trig identity — the cleanest path is to introduce that parameter explicitly and use $\dfrac{dy}{dx}=\dfrac{dy/du}{dx/du}$, rather than trying to eliminate $t$ directly.

   Step-by-Step Solution

   1. For $|t|\ge1$, the standard identity $\operatorname{cosec}^{-1}t+\sec^{-1}t=\dfrac\pi2$ holds.
   2. Let $u=\operatorname{cosec}^{-1}t$. Then $\sec^{-1}t=\dfrac\pi2-u$.
   3. $x=\sqrt{2^u}=2^{u/2}$, and $y=\sqrt{2^{\pi/2-u}}=2^{(\pi/2-u)/2}=2^{\pi/4-u/2}$.
   4. Differentiate w.r.t. $u$: $\dfrac{dx}{du}=2^{u/2}\ln2\cdot\dfrac12=\dfrac{x\ln2}2$.
   5. $\dfrac{dy}{du}=2^{\pi/4-u/2}\ln2\cdot\left(-\dfrac12\right)=-\dfrac{y\ln2}2$. …

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10. 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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11. 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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12. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.If $x = 2\cos^3\theta$ and $y = 3\sin^2\theta$, then $\frac{dy}{dx} =$ (A) $-\sec\theta$ (B) $\cos\theta$ (C) $-\mathrm{cosec}\,\theta$ (D) $\sin\theta$

    ›Reveal solutionSolution

    Differentiate $x$ and $y$ separately with respect to the parameter $\theta$, then take the ratio. Answer: $-\sec\theta$.

    Concept and Intuition

    For a parametric curve $x=x(\theta),\ y=y(\theta)$, the derivative is found via the chain rule as $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$, avoiding the need to eliminate $\theta$ and differentiate implicitly.

    Step-by-Step Solution

    1. $x=2\cos^3\theta$. Differentiate: $\dfrac{dx}{d\theta}=2\cdot3\cos^2\theta\cdot(-\sin\theta)=-6\cos^2\theta\sin\theta$.
    2. $y=3\sin^2\theta$. Differentiate: $\dfrac{dy}{d\theta}=3\cdot2\sin\theta\cos\theta=6\sin\theta\cos\theta$.
    3. $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{6\sin\theta\cos\theta}{-6\cos^2\theta\sin\theta}$. …

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