Q.Find $\frac{dy}{dx}$ in the following: $e^{6x} \cos 3x$

Showing the 12 most recent of 23 on this concept.

1. 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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2. 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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3. 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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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-2021-08-24-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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6. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $y = \sin(\sin x)$ and $y'' + f(x)\cdot y' + g(x)\cdot y = 0$, then $f(x)\cdot g(x) =$ ______ (A) $\dfrac{1}{2}\sin(2x)$ (B) $\dfrac{1}{2}\cos(2x)$ (C) $\sin(2x)$ (D) $\cos(2x)$

   ›Reveal solutionSolution

   This tests forming the second-order ODE satisfied by $y=\sin(\sin x)$ by eliminating trig functions in favor of $y$ and $y'$. The answer is $f(x)g(x)=\tfrac12\sin2x$.

   Concept and Intuition

   The idea is to differentiate $y$ twice, then rewrite the resulting expression purely in terms of $y$ and $y'$ (using the fact that $y'$ itself contains $\cos(\sin x)$), so that we can read off $f(x)$ and $g(x)$ by matching coefficients.

   Step-by-Step Solution

   1. $y=\sin(\sin x)\Rightarrow y'=\cos x\cos(\sin x)$.
   2. $y''=-\sin x\cos(\sin x)+\cos x\cdot(-\sin(\sin x))\cos x=-\sin x\cos(\sin x)-\cos^2x\sin(\sin x)$.
   3. From step 1, $\cos(\sin x)=\dfrac{y'}{\cos x}$, so $-\sin x\cos(\sin x)=-\sin x\cdot\dfrac{y'}{\cos x}=-\tan x\,y'$.
   4. Also $\sin(\sin x)=y$, so the second term is $-\cos^2x\,y$.
   5. Hence $y''=-\tan x\,y'-\cos^2x\,y$, i.e. $y''+\tan x\,y'+\cos^2x\,y=0$. …

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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 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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9. 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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10. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.If $y = \sin\left(2\tan^{-1}\left(\sqrt{\dfrac{1-x}{1+x}}\right)\right)$, $x = \cos 2\theta$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{x}{\sqrt{1 - x^2}}$ (B) $-\cot 2\theta$ (C) $\tan 2\theta$ (D) $\dfrac{-x}{2\sqrt{1 - x^2}}$

    ›Reveal solutionSolution

    Substituting $x=\cos2\theta$ collapses the arctan expression to $\theta$ itself, turning this into a simple parametric-derivative problem.

    Concept and Intuition

    $\sqrt{\dfrac{1-\cos2\theta}{1+\cos2\theta}}=\sqrt{\dfrac{2\sin^2\theta}{2\cos^2\theta}}=|\tan\theta|$, and $2\tan^{-1}(\tan\theta)=2\theta$ (for $\theta$ in the principal range), so $y=\sin2\theta$ directly — the whole problem becomes parametric differentiation with parameter $\theta$.

    Step-by-Step Solution

    1. $\dfrac{1-\cos2\theta}{1+\cos2\theta}=\dfrac{2\sin^2\theta}{2\cos^2\theta}=\tan^2\theta\Rightarrow\sqrt{\cdots}=\tan\theta$.
    2. $\tan^{-1}(\tan\theta)=\theta\Rightarrow y=\sin(2\theta)$.
    3. Parametrically, $x=\cos2\theta$, $y=\sin2\theta$: $\dfrac{dx}{d\theta}=-2\sin2\theta$, $\dfrac{dy}{d\theta}=2\cos2\theta$. …

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11. 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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12. 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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