Q.Find $\frac{dy}{dx}$ in the following: $\sqrt{e^{\sqrt{x}}}, x > 0$

Showing the 12 most recent of 23 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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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 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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11. 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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12. 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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