Q.Find $\frac{dy}{dx}$ in the following: $\sin (ax + b)$

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 $2y\,y'=\dfrac{\sinh x\,(2\sqrt{\cosh x}+1)}{2\sqrt{\cosh x}}$.
   5. Solve for $y'$: $y'=\dfrac{\sinh x\,(1+2\sqrt{\cosh x})}{4y\sqrt{\cosh x}}$.

   Common Mistakes

   • Differentiating the nested radical directly without squaring first, which tangles two chain-rule layers and often drops a factor of 2.
   • Forgetting to carry the $2y$ from the implicit differentiation (i.e., leaving the answer in terms of $y^2$ instead of solving for $y'$ explicitly).

   ✓Final answer

   The correct option is (D) — $\dfrac{\sinh x(1+2\sqrt{\cosh x})}{4y\sqrt{\cosh x}}$.

   ANSWER: D

2. 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. $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}=\dfrac{6\sin\theta\cos\theta}{-6\cos^2\theta\sin\theta}$.
   4. Cancel the common factor $6\sin\theta\cos\theta$ (assuming $\sin\theta,\cos\theta\ne0$): $=\dfrac{1}{-\cos\theta}=-\sec\theta$.

   Common Mistakes

   • Forgetting the chain-rule factor when differentiating $\cos^3\theta$ or $\sin^2\theta$ (missing the "3" or "2" power-rule multiplier).
   • Sign error when cancelling $\sin\theta\cos\theta$ from numerator and denominator, dropping the leading minus sign.

   ✓Final answer

   The correct option is (A) — $-\sec\theta$.

   ANSWER: A

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. $y''+\tan x\,y'+\cos^2x\,y=0$.
   6. Matching to $y''+f(x)y'+g(x)y=0$: $f(x)=\tan x$, $g(x)=\cos^2x$.
   7. $f(x)g(x)=\tan x\cos^2x=\sin x\cos x=\tfrac12\sin(2x)$.

   Common Mistakes

   • Leaving the answer in terms of $\sin(\sin x)$ and $\cos(\sin x)$ instead of eliminating them in favor of $y,y'$.
   • Sign slip when differentiating $-\cos(\sin x)\cdot\sin x$ the second time.

   ✓Final answer

   The correct option is (A) — $\dfrac12\sin(2x)$.

   ANSWER: A

4. 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, $x=\cos2\theta$, $y=\sin2\theta$: $\dfrac{dx}{d\theta}=-2\sin2\theta$, $\dfrac{dy}{d\theta}=2\cos2\theta$.
   4. $\dfrac{dy}{dx}=\dfrac{2\cos2\theta}{-2\sin2\theta}=-\cot2\theta$.

   Common Mistakes

   • Forgetting the $|\cdot|$/domain subtlety in $\sqrt{\tan^2\theta}=|\tan\theta|$ (fine within the principal branch used here).
   • Differentiating $y$ directly w.r.t. $x$ via chain rule without simplifying first — it's algebraically messier and error-prone.

   ✓Final answer

   The correct option is (B) — $-\cot 2\theta$.

   ANSWER: B

5. 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 $\dfrac{2Au'}{-x^3} = \dfrac{-2A}{x^3}\cdot\dfrac{-3x^2}{2u} = \dfrac{3A}{xu}$.
   5. This must equal $\dfrac{1}{x\sqrt{1-x^3}} = \dfrac{1}{xu}$, so $3A=1 \Rightarrow A=\dfrac13$.
   6. Hence $AB = \dfrac13\times(-1) = -\dfrac13$.

   Common Mistakes

   • Trying random values of $B$ without checking which one makes $(u+B)(u+1)$ collapse to a pure power of $x$ — $B=-1$ is the key insight (difference of squares).
   • Sign slip in $u'$ (the exponent's chain rule brings in a negative sign from $-x^3$'s derivative).

   ✓Final answer

   The correct option is (B) — $\dfrac{-1}{3}$.

   ANSWER: B

6. 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. $\dfrac{dy}{dx}=\dfrac{6(\csc t-\sin t)}{3(\cos t+\csc t)}=\dfrac{2(\csc t-\sin t)}{\cos t+\csc t}$.
   5. Simplify: $\csc t-\sin t=\dfrac{1-\sin^2t}{\sin t}=\dfrac{\cos^2t}{\sin t}$; and $\cos t+\csc t=\dfrac{\sin t\cos t+1}{\sin t}$.
   6. So $\dfrac{dy}{dx}=\dfrac{2\cos^2t/\sin t}{(1+\sin t\cos t)/\sin t}=\dfrac{2\cos^2t}{1+\sin t\cos t}$.

   Common Mistakes

   • Misremembering the sign in $\dfrac{d}{dt}\log\cot(t/2)$ (it is $-\csc t$, not $+\csc t$).
   • Not cancelling the common $\sin t$ factor from numerator and denominator, leaving an unsimplified answer that doesn't match any option.

   ✓Final answer

   The correct option is (C) — $\dfrac{2\cos^2 t}{1+\sin t\cos t}$.

   ANSWER: C

7. 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: $\dfrac{d}{dx}\sec(x/2) = \sec(x/2)\tan(x/2)\cdot\dfrac12$.
   6. Comparing to $k\sec(x/2)\tan(x/2)$: $k=\dfrac12$.

   Common Mistakes

   • Missing the chain rule factor of $\tfrac12$ from differentiating $\sec(x/2)$ (the "$x/2$" inside contributes an extra $\tfrac12$).
   • Sign errors taking square roots (assuming the wrong sign of $\cos$ at some stage), though this doesn't affect the magnitude of $k$ for the expected range.

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{2}$.

   ANSWER: A

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: $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}=\dfrac{1+x}{2x}\cdot\left(\dfrac{-1}{u(1+x)^2}\right)=\dfrac{-1}{2x\,u(1+x)}$.
   5. Simplify $u(1+x)$: $u(1+x)=\sqrt{\dfrac{1-x}{1+x}}\cdot(1+x)=\sqrt{(1-x)(1+x)}=\sqrt{1-x^2}$.
   6. So $\dfrac{dy}{dx}=\dfrac{-1}{2x\sqrt{1-x^2}}$.

   Common Mistakes

   • Forgetting the extra factor of $x$ that survives from $\dfrac{1}{1-u^2}=\dfrac{1+x}{2x}$, leading to option (A)'s answer (missing the $x$ in the denominator).
   • Sign errors differentiating $\dfrac{1-x}{1+x}$ via the quotient rule.

   ✓Final answer

   The correct option is (B) — $-\dfrac{1}{2x\sqrt{1-x^2}}$.

   ANSWER: B

9. 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 $|u|=u=\dfrac{1+x^2}{2x}$.
   5. $u\sqrt{u^2-1} = \dfrac{1+x^2}{2x}\cdot\dfrac{x^2-1}{2x} = \dfrac{(1+x^2)(x^2-1)}{4x^2}$.
   6. $\dfrac{dy}{dx} = \dfrac{u'}{u\sqrt{u^2-1}} = \dfrac{(x^2-1)/(2x^2)}{(1+x^2)(x^2-1)/(4x^2)} = \dfrac{x^2-1}{2x^2}\times\dfrac{4x^2}{(1+x^2)(x^2-1)} = \dfrac{4}{2(1+x^2)} = \dfrac{2}{1+x^2}$.

   Common Mistakes

   • Dropping the absolute value carelessly without checking the sign of $x$ in the given domain $x>1$.
   • Forgetting to simplify $u^2-1$ using the difference-of-squares-of-squares factoring trick, leading to a messy, harder-to-simplify square root.

   ✓Final answer

   The correct option is (B) — $\dfrac{2}{1+x^2}$.

   ANSWER: B

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).
    4. Add: $\dfrac{dy}{dx}=\dfrac{1}{x\sqrt{x^2-1}}+\dfrac{1}{\sqrt{x^2-1}}=\dfrac{1+x}{x\sqrt{x^2-1}}$.

    Common Mistakes

    • Using $\sqrt{x^2}=|x|$ and leaving an unnecessary absolute value, when the domain $x>1$ already fixes the sign.
    • Misremembering the derivative of $\mathrm{Sinh}^{-1}u$ as $\dfrac{1}{1+u^2}$ (that's $\mathrm{Tan}^{-1}$'s formula) instead of $\dfrac{1}{\sqrt{1+u^2}}$.

    ✓Final answer

    The correct option is (B) — $\dfrac{x+1}{x\sqrt{x^2-1}}$.

    ANSWER: B

11. 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. $u\sqrt{1-u^2}=\dfrac{9}{9x^2+10}\cdot\dfrac{\sqrt{(9x^2+1)(9x^2+19)}}{9x^2+10}=\dfrac{9\sqrt{(9x^2+1)(9x^2+19)}}{(9x^2+10)^2}$.
    6. $\dfrac{dy}{dx}=\dfrac{-1}{u\sqrt{1-u^2}}\cdot\dfrac{du}{dx}=\dfrac{-(9x^2+10)^2}{9\sqrt{(9x^2+1)(9x^2+19)}}\cdot\dfrac{-162x}{(9x^2+10)^2}=\dfrac{162x}{9\sqrt{(9x^2+1)(9x^2+19)}}=\dfrac{18x}{\sqrt{(9x^2+1)(9x^2+19)}}$.

    Common Mistakes

    • Missing the sign cancellation between the $-1$ in the $\operatorname{sech}^{-1}$ derivative formula and the negative $du/dx$, which would flip the final sign.
    • Not recognizing $(9x^2+10)^2-81$ as a factorable difference of squares and instead leaving it unsimplified (masking the match with the given options).

    ✓Final answer

    The correct option is (C) — $\dfrac{18x}{\sqrt{(9x^2+19)(9x^2+1)}}$.

    ANSWER: C

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. $\dfrac{dy}{du}=2^{\pi/4-u/2}\ln2\cdot\left(-\dfrac12\right)=-\dfrac{y\ln2}2$.
    6. $\dfrac{dy}{dx}=\dfrac{dy/du}{dx/du}=\dfrac{-y\ln2/2}{x\ln2/2}=-\dfrac yx$.

    Common Mistakes

    • Forgetting the identity linking $\operatorname{cosec}^{-1}t$ and $\sec^{-1}t$ and instead trying to differentiate $x$ and $y$ directly with respect to $t$ using the (more complicated) derivative of $\operatorname{cosec}^{-1}t$.
    • Sign error: missing the minus sign that comes from $\sec^{-1}t=\pi/2-u$ (a decreasing function of $u$).

    ✓Final answer

    The correct option is (C) — $-\dfrac{y}{x}$.

    ANSWER: C