Q.Differentiate w.r.t. $x$: $\sin x^2 + \sin^2 x + \sin^2(x^2)$.

Showing the 12 most recent of 23 on this concept.

1. 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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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 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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4. 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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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $y = \mathrm{Sec}^{-1}\left(\dfrac{1+x^2}{2x}\right)$ and $x > 1$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{1}{1+x^2}$ (B) $\dfrac{2}{1+x^2}$ (C) $-\dfrac{1}{1+x^2}$ (D) $-\dfrac{2}{1+x^2}$

   ›Reveal solutionSolution

   Differentiating the inverse secant of a rational expression using the chain rule and careful algebraic simplification gives $\dfrac{dy}{dx}=\dfrac{2}{1+x^2}$.

   Concept and Intuition

   For $y=\mathrm{Sec}^{-1}(u)$, the derivative formula is $\dfrac{dy}{dx} = \dfrac{1}{|u|\sqrt{u^2-1}}\cdot\dfrac{du}{dx}$. Here $u=\dfrac{1+x^2}{2x}$ is positive for $x>1$, so we can drop the absolute value and just carefully simplify the algebra — the key insight is recognizing that $u^2-1$ factors as a perfect square-like expression involving $(x^2-1)^2$, which simplifies the square root beautifully.

   Step-by-Step Solution

   1. Let $u = \dfrac{1+x^2}{2x}$. Compute $\dfrac{du}{dx}$ using the quotient rule: $u' = \dfrac{2x(2x) - (1+x^2)(2)}{(2x)^2} = \dfrac{4x^2-2-2x^2}{4x^2} = \dfrac{2x^2-2}{4x^2} = \dfrac{x^2-1}{2x^2}$.
   2. Compute $u^2-1$: $u^2-1 = \dfrac{(1+x^2)^2-4x^2}{4x^2} = \dfrac{(1+x^2-2x)(1+x^2+2x)}{4x^2} = \dfrac{(x-1)^2(x+1)^2}{4x^2}$.
   3. So $\sqrt{u^2-1} = \dfrac{|x-1||x+1|}{2|x|} = \dfrac{|x^2-1|}{2|x|}$. For $x>1$: $x^2-1>0$ and $x>0$, so $\sqrt{u^2-1}=\dfrac{x^2-1}{2x}$.
   4. For $x>1$, $u = \dfrac{1+x^2}{2x}>0$, so $|u|=u=\dfrac{1+x^2}{2x}$. …

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6. 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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7. 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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8. 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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9. 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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10. 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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11. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
    Q.Find the value of 'k' if $\dfrac{d}{dx}\left\{\dfrac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2\cos(4x)}}}}\right\} = k\sec\left(\dfrac{x}{2}\right)\tan\left(\dfrac{x}{2}\right)$ (A) $\dfrac{1}{2}$ (B) 2 (C) 1 (D) $\dfrac{1}{8}$

    ›Reveal solutionSolution

    Repeatedly apply the half-angle identity $2+2\cos\phi=4\cos^2(\phi/2)$ to peel away the nested square roots, collapsing the whole expression to a simple $\sec(x/2)$.

    Concept and Intuition

    The identity $1+\cos\phi=2\cos^2(\phi/2)$ (i.e. $2+2\cos\phi=4\cos^2(\phi/2)$) is exactly designed to simplify nested square-root expressions of this kind — applying it repeatedly, from the innermost root outward, collapses the whole tower.

    Step-by-Step Solution

    1. Innermost: $2+2\cos4x = 4\cos^2(2x)$ (using $2+2\cos\phi=4\cos^2(\phi/2)$ with $\phi=4x$). So $\sqrt{2+2\cos4x}=2\cos2x$ (taking $\cos2x>0$).
    2. Next level: $2+\sqrt{2+2\cos4x}=2+2\cos2x = 4\cos^2x$. So $\sqrt{2+\sqrt{2+2\cos4x}} = 2\cos x$.
    3. Next level: $2+\sqrt{2+\sqrt{2+2\cos4x}} = 2+2\cos x = 4\cos^2(x/2)$. So $\sqrt{2+\sqrt{2+\sqrt{2+2\cos4x}}} = 2\cos(x/2)$.
    4. So the whole expression is $\dfrac{2}{2\cos(x/2)} = \sec(x/2)$.
    5. Differentiate: $\dfrac{d}{dx}\sec(x/2) = \sec(x/2)\tan(x/2)\cdot\dfrac12$. …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.If $f(x) = \sec^{-1}\left(\dfrac{1}{2x^2-1}\right)$ and $g(x) = \tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)$, then the derivative of $f(x)$ with respect to $g(x)$ is (A) $\dfrac{1+x^2}{4\sqrt{1-x^2}}$ (B) $\dfrac{1-x^2}{4\sqrt{1+x^2}}$ (C) $-\dfrac{4(1-x^2)}{\sqrt{1+x^2}}$ (D) $-\dfrac{4(1+x^2)}{\sqrt{1-x^2}}$

    ›Reveal solutionSolution

    Simplify both $f$ and $g$ using standard inverse-trig substitutions, then take the ratio of derivatives. Answer: $\dfrac{df}{dg}=-\dfrac{4(1+x^2)}{\sqrt{1-x^2}}$.

    Concept and Intuition

    $\dfrac{df}{dg}$ means $\dfrac{df/dx}{dg/dx}$. Both $f$ and $g$ are compositions of inverse trig functions that simplify beautifully with the right substitution ($x=\cos\theta$-type for $f$, $x=\tan\theta$ for $g$), turning ugly inverse-trig expressions into simple linear multiples of $\tan^{-1}x$ or $\cos^{-1}x$.

    Step-by-Step Solution

    1. $f(x)=\sec^{-1}\left(\dfrac{1}{2x^2-1}\right)=\cos^{-1}(2x^2-1)$ since $\sec^{-1}(1/u)=\cos^{-1}u$.
    2. For $x\in(0,1)$, write $x=\cos\theta$: then $2x^2-1=2\cos^2\theta-1=\cos2\theta$, so $f=\cos^{-1}(\cos2\theta)=2\theta=2\cos^{-1}x$.
    3. $f'(x) = 2\cdot\left(-\dfrac{1}{\sqrt{1-x^2}}\right) = -\dfrac{2}{\sqrt{1-x^2}}$.
    4. For $g$: put $x=\tan\theta$, so $\sqrt{1+x^2}=\sec\theta$. Then $\dfrac{\sqrt{1+x^2}-1}{x}=\dfrac{\sec\theta-1}{\tan\theta}=\dfrac{1-\cos\theta}{\sin\theta}=\tan\left(\dfrac{\theta}{2}\right)$.
    5. So $g(x)=\tan^{-1}\left(\tan\dfrac{\theta}{2}\right)=\dfrac{\theta}{2}=\dfrac12\tan^{-1}x$. …

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