Q.Find $\frac{dy}{dx}$ in the following: $y = \sec^{-1} \left(\frac{1}{2x^2-1}\right), 0 < x < \frac{1}{\sqrt{2}}$

Showing the 12 most recent of 13 on this concept.

1. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.If $y = \sin\left(\tan^{-1}\left(\dfrac{1}{\sqrt{x^2 - 1}}\right)\right),\ x > 1$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{1}{x^2}$ (B) $\dfrac{1}{x^4}$ (C) $\dfrac{-1}{x^2}$ (D) $\dfrac{-1}{x^4}$ (E) $\dfrac{1}{x^3}$

   ›Reveal solutionSolution

   Simplify the inverse trig: the angle whose tangent is $\dfrac{1}{\sqrt{x^2-1}}$ has $\sin=\dfrac1x$, so $y=\dfrac1x$ and its derivative is $-\dfrac{1}{x^2}$.

   Let $\theta=\tan^{-1}\!\left(\dfrac{1}{\sqrt{x^2-1}}\right)$, so $\tan\theta=\dfrac{1}{\sqrt{x^2-1}}$ with opposite $=1$ and adjacent $=\sqrt{x^2-1}$. The hypotenuse is $\sqrt{1+(x^2-1)}=\sqrt{x^2}=x$ (since $x>1$) …

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2. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.If $s = \sqrt{t + 1},\ x = \log s$ and $y = 6x + 3$, then $\dfrac{dy}{dt} =$ (A) $\dfrac{2}{\sqrt{t + 1}}$ (B) $\dfrac{6}{t + 1}$ (C) $3\sqrt{t + 1}$ (D) $\dfrac{3}{t + 1}$ (E) $\dfrac{3}{\sqrt{t + 1}}$

   ›Reveal solutionSolution

   Substituting back, $y=6\log\sqrt{t+1}+3=3\log(t+1)+3$, whose $t$-derivative is $\dfrac{3}{t+1}$.

   With $s=\sqrt{t+1}$ and $x=\log s$, we have $x=\log\sqrt{t+1}=\tfrac12\log(t+1)$. Then

   $y=6x+3=6\cdot\tfrac12\log(t+1)+3=3\log(t+1)+3.$ …

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3. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.If $x = \sec\theta - \cos\theta$, $y = \sec^{10}\theta - \cos^{10}\theta$, then $\left(\frac{dy}{dx}\right)^2$ is equal to (A) $100\left(\frac{y^2 + 4}{x^2 + 4}\right)$ (B) $100\left(\frac{y^4 - 4}{x^4 + 4}\right)$ (C) $100\left(\frac{y^2 + 4}{x^2 - 4}\right)$ (D) $100\left(\frac{y^4 + 2}{x^4 + 4}\right)$ (E) $100\left(\frac{y^4 + 4}{x^4 + 2}\right)$

   ›Reveal solutionSolution

   The key identities $x^2+4=(\sec\theta+\cos\theta)^2$ and $y^2+4=(\sec^{10}\theta+\cos^{10}\theta)^2$ turn $(dy/dx)^2$ into a clean ratio.

   Since $x=\sec\theta-\cos\theta$, $x^2+4=\sec^2\theta+\cos^2\theta+2=(\sec\theta+\cos\theta)^2$.

   Since $y=\sec^{10}\theta-\cos^{10}\theta$, $y^2+4=\sec^{20}\theta+\cos^{20}\theta+2=(\sec^{10}\theta+\cos^{10}\theta)^2$. …

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4. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.Let $h(x) = f\left(\sqrt{g(x)}\right)$. If $f'(3) = 6$, $g'(3) = 3$ and $g(3) = 9$, then the value of $h'(3)$ is equal to (A) $1$ (B) $3$ (C) $6$ (D) $9$ (E) $18$

   ›Reveal solutionSolution

   By the chain rule h'(3) = f'(3)g'(3)/(2sqrt(g(3))) = 6*3/6 = 3.

   Concept and Intuition

   h(x) = f(sqrt(g(x))) is a triple composition; differentiate outer-to-inner, picking up the derivative of the square root and of g.

   Step-by-Step Solution

   1. h'(x) = f'(sqrt(g(x))) * d/dx[sqrt(g(x))] = f'(sqrt(g)) * g'(x)/(2*sqrt(g(x))).
   2. At x = 3: g(3) = 9 so sqrt(g) = 3, f'(3) = 6, g'(3) = 3. …

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5. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.If $y = \tan^{-1}(x^2 - x)$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{2x}{1 + (x^2 - x)^2}$ (B) $\dfrac{2x - 1}{1 + (x^2 - x)^2}$ (C) $\dfrac{2x - 1}{1 - (x^2 - x)^2}$ (D) $\dfrac{-2x + 1}{1 + (x^2 - x)^2}$ (E) $(2x - 1)\left(1 + (x^2 - x)^2\right)$

   ›Reveal solutionSolution

   d/dx tan^{-1}(u) = u'/(1+u^2) with u=x^2-x gives (2x-1)/(1+(x^2-x)^2).

   Concept and Intuition

   The derivative of arctan(u) is u'/(1+u^2). Here u = x^2 - x so u' = 2x - 1.

   Step-by-Step Solution

   1. Let u = x^2 - x, so u' = 2x - 1.
   2. dy/dx = u'/(1+u^2) = (2x-1)/(1+(x^2-x)^2).

   Common Mistakes …

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6. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.For $x \in \mathbb{R}$, let $f(x) = \log 3 - \sin x$ and $g(x) = f(f(x))$. Then $g'(0) =$ (A) $\sin(\log 3)$ (B) $-\sin(\log 3)$ (C) $-\cos(\log 3)$ (D) $2\cos(\log 3)$ (E) $\cos(\log 3)$

   ›Reveal solutionSolution

   With $f(x)=\log3-\sin x$, $f'(x)=-\cos x$; by the chain rule $g'(0)=f'(f(0))f'(0)=(-\cos(\log3))(-\cos0)=\cos(\log3)$.

   Here $f(x)=\log3-\sin x$ so $f'(x)=-\cos x$. Then $f(0)=\log3-\sin0=\log3$ and $f'(0)=-\cos0=-1$. …

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7. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.If $u = \sec^{-1}(-\sec 2\theta)$ and $v = \cos\theta$, then $\dfrac{du}{dv}$ at $\theta = \dfrac{\pi}{4}$, is equal to (A) $\sqrt{2}$ (B) $2\sqrt{2}$ (C) $\dfrac{1}{\sqrt{2}}$ (D) $\dfrac{1}{2\sqrt{2}}$ (E) $-\sqrt{2}$

   ›Reveal solutionSolution

   Simplify $u=\pi-2\theta$, differentiate both $u$ and $v$ in $\theta$, divide.

   Using $\sec^{-1}(-x)=\pi-\sec^{-1}(x)$ and $\sec^{-1}(\sec 2\theta)=2\theta$ (for $2\theta$ in the principal range),

   $u=\sec^{-1}(-\sec 2\theta)=\pi-2\theta\;\Rightarrow\;\frac{du}{d\theta}=-2.$

   With $v=\cos\theta$, $\dfrac{dv}{d\theta}=-\sin\theta$. Hence …

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8. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.If $y = \sin x \sin 2x$, and $t = \cos x$, then $\frac{dy}{dt}$ is (A) $2(3t^2 - 1)$ (B) $1 - 3t^2$ (C) $\frac{1}{2}(1 - 3t^2)$ (D) $(3t^2 - 1)$ (E) $2(1 - 3t^2)$

   ›Reveal solutionSolution

   Express $y$ in $t=\cos x$: $y=2t-2t^3$, then $\dfrac{dy}{dt}=2-6t^2=2(1-3t^2)$.

   With $t=\cos x$ and using $\sin 2x=2\sin x\cos x$:

   $y=\sin x\,\sin 2x=\sin x\,(2\sin x\cos x)=2\sin^2x\cos x.$

   Since $\sin^2x=1-\cos^2x=1-t^2$ and $\cos x=t$, …

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9. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.If $x^3 = \sin\theta$, $y^3 = \cos\theta$, then $x\frac{dy}{dx}$ is (A) $\frac{y^5 - 1}{y^5}$ (B) $\frac{y^6 - 1}{y^5}$ (C) $\frac{y^6 - 1}{y^6}$ (D) $\frac{y^3 - 1}{y^3}$ (E) $\frac{y^2 - 1}{y^2}$

   ›Reveal solutionSolution

   Differentiate both parametric relations with respect to $\theta$, form $\dfrac{dy}{dx}$, and substitute $x^6=1-y^6$.

   Concept. Here $x$ and $y$ are both given as functions of a parameter $\theta$. For parametric curves, $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$.

   Step 1 — differentiate w.r.t. $\theta$.

   $x^3=\sin\theta\ \Rightarrow\ 3x^2\frac{dx}{d\theta}=\cos\theta,\qquad y^3=\cos\theta\ \Rightarrow\ 3y^2\frac{dy}{d\theta}=-\sin\theta.$

   Step 2 — form $\dfrac{dy}{dx}$.

   $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{-\sin\theta/(3y^2)}{\cos\theta/(3x^2)}=-\frac{x^2\sin\theta}{y^2\cos\theta}.$

   Since $\sin\theta=x^3$ and $\cos\theta=y^3$, …

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10. KEAM 2024Set eng-2024-06074 marksMCQ
    Q.If $y=\log_e\left(\dfrac{1+2x^2}{1-3x^2}\right)$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{10x}{1-x^2-6x^4}$ (B) $\dfrac{12x^3}{1-x^2-6x^4}$ (C) $\dfrac{10x}{1-6x^4}$ (D) $\dfrac{-10x}{1-x^2-6x^4}$ (E) $\dfrac{-12x^3}{1-x^2-6x^4}$

    ›Reveal solutionSolution

    Split the log, differentiate each term, combine over the common denominator.

    $y=\log(1+2x^2)-\log(1-3x^2).$

    Differentiating:

    $\frac{dy}{dx}=\frac{4x}{1+2x^2}-\frac{-6x}{1-3x^2}=\frac{4x}{1+2x^2}+\frac{6x}{1-3x^2}.$

    Common denominator $(1+2x^2)(1-3x^2)=1-x^2-6x^4$; numerator: …

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11. KEAM 2024Set eng-2024-06084 marksMCQ
    Q.The derivative of $t^2+t$ with respect to $t-1$ at $t=-2$, is equal to (A) $-4$ (B) $2$ (C) $-1$ (D) $-3$ (E) $-\dfrac{1}{2}$

    ›Reveal solutionSolution

    Differentiate parametrically: divide $\dfrac{d(t^2+t)}{dt}$ by $\dfrac{d(t-1)}{dt}$, then substitute $t=-2$.

    Let $u=t^2+t$ and $w=t-1$. The derivative of $u$ with respect to $w$ is

    $\frac{du}{dw}=\frac{du/dt}{dw/dt}.$ …

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12. KEAM 2022Set eng-2022-P2-B14 marksMCQ
    Q.If $y = e^{3\log(2x+1)}$, then $\frac{dy}{dx} =$ (A) $6e^{3\log(2x+1)}$ (B) $6\frac{e^{3\log(2x+1)}}{2x+1}$ (C) $\frac{e^{3\log(2x+1)}}{2x+1}$ (D) $\frac{e^{3\log(2x+1)}}{3(2x+1)}$ (E) $(2x+1)e^{3\log(2x+1)}$

    ›Reveal solutionSolution

    $\frac{dy}{dx}=\frac{6\,e^{3\log(2x+1)}}{2x+1}$.

    Concept and Intuition

    Differentiate the exponential by the chain rule: $\frac{d}{dx}e^{u}=e^{u}u'$, with $u=3\log(2x+1)$.

    Step-by-Step Solution

    1. $u=3\log(2x+1)$, so $u'=3\cdot\frac{2}{2x+1}=\frac{6}{2x+1}$.
    2. $\frac{dy}{dx}=e^{3\log(2x+1)}\cdot u'=e^{3\log(2x+1)}\cdot\frac{6}{2x+1}$.
    3. Hence $\frac{dy}{dx}=\frac{6\,e^{3\log(2x+1)}}{2x+1}$.

    Common Mistakes …

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