Q.Differentiate w.r.t. $x$, the following function:

Showing the 12 most recent of 13 on this concept.

1. 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:

   $4x(1-3x^2)+6x(1+2x^2)=4x-12x^3+6x+12x^3=10x.$

   So $\frac{dy}{dx}=\dfrac{10x}{1-x^2-6x^4}$.

   ✓Final answer

   The correct option is (A).

2. 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

   • Forgetting the inner factor $2$ from $\frac{d}{dx}(2x+1)$.
   • Omitting the $\frac{1}{2x+1}$ derivative of the log.

   ✓Final answer

   The correct option is (B) — $6\frac{e^{3\log(2x+1)}}{2x+1}$.

   ANSWER: B

3. 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

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

   Differentiating with respect to $t$: $\dfrac{dy}{dt}=3\cdot\dfrac{1}{t+1}=\dfrac{3}{t+1}$.

   ✓Final answer

   The correct option is (D).

4. 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 $x>1$). Hence $\sin\theta=\dfrac{1}{x}$, i.e. $y=\dfrac1x=x^{-1}$.

   Differentiating, $\dfrac{dy}{dx}=-x^{-2}=-\dfrac{1}{x^2}$.

   ✓Final answer

   The correct option is (C).

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

   Differentiating and forming $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$ leads to $\left(\dfrac{dy}{dx}\right)^2=100\,\dfrac{(\sec^{10}\theta+\cos^{10}\theta)^2}{(\sec\theta+\cos\theta)^2}=100\left(\dfrac{y^2+4}{x^2+4}\right)$.

   A numerical check at $\theta=\frac{\pi}{3}$ confirms this (both sides $\approx 1.678\times10^7$).

   ✓Final answer

   The correct option is (A).

6. 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

   • Writing the denominator as 1 - u^2 (that belongs to artanh, not arctan).

   ✓Final answer

   The correct option is (B) — (2x-1)/(1+(x^2-x)^2).

   ANSWER: B

7. 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.
   3. h'(3) = 6 * 3/(2*3) = 6 * 3/6 = 3.

   Common Mistakes

   • Forgetting the 1/(2*sqrt(g)) factor from the square root.

   ✓Final answer

   The correct option is (B) — 3.

   ANSWER: B

8. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.If $f(x) = \left(x^3 + \sin \pi x\right)^5$, then $f'(1)$ is equal to (A) $2^5$ (B) $5(2^4)$ (C) $15$ (D) $5(3 + \pi)$ (E) $5(3 - \pi)$

   ›Reveal solutionSolution

   $f'(1)=5(3-\pi)$.

   Concept and Intuition

   Differentiate the outer fifth power via the chain rule and multiply by the derivative of the inner expression, then evaluate at $x=1$.

   Step-by-Step Solution

   1. $f'(x)=5(x^3+\sin\pi x)^4\cdot(3x^2+\pi\cos\pi x)$.
   2. At $x=1$: inner base $=1+\sin\pi=1$, so $1^4=1$.
   3. Inner derivative $=3(1)+\pi\cos\pi=3-\pi$.
   4. $f'(1)=5\cdot1\cdot(3-\pi)=5(3-\pi)$.

   Common Mistakes

   • Using $\cos\pi=+1$ (it is $-1$), which flips the sign to $5(3+\pi)$.

   ✓Final answer

   The correct option is (E) — $5(3-\pi)$.

   ANSWER: E

9. 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

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

   Compute each piece: $\dfrac{du}{dt}=2t+1$ and $\dfrac{dw}{dt}=1$.

   Hence $\dfrac{du}{dw}=\dfrac{2t+1}{1}=2t+1$.

   At $t=-2$: $2(-2)+1=-4+1=-3$.

   ✓Final answer

   The correct option is (D).

10. 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 $f'(0)=-\cos0=-1$.

    For $g(x)=f(f(x))$, the chain rule gives $g'(x)=f'(f(x))\,f'(x)$. At $x=0$:

    $g'(0)=f'(\log3)\cdot f'(0)=(-\cos(\log3))\cdot(-1)=\cos(\log3).$

    ✓Final answer

    The correct option is (E).

11. 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

    $\frac{du}{dv}=\frac{du/d\theta}{dv/d\theta}=\frac{-2}{-\sin\theta}=\frac{2}{\sin\theta}.$

    At $\theta=\dfrac{\pi}{4}$, $\sin\theta=\dfrac{1}{\sqrt2}$, so $\dfrac{du}{dv}=2\sqrt2$.

    ✓Final answer

    The correct option is (B).

12. 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 $\cos\theta=y^3$,

    $\frac{dy}{dx}=-\frac{x^2\cdot x^3}{y^2\cdot y^3}=-\frac{x^5}{y^5}.$

    Step 3 — multiply by $x$.

    $x\frac{dy}{dx}=-\frac{x^6}{y^5}.$

    Step 4 — eliminate $x$ using the Pythagorean identity.

    $x^6=(x^3)^2=\sin^2\theta=1-\cos^2\theta=1-(y^3)^2=1-y^6.$

    Hence

    $x\frac{dy}{dx}=-\frac{1-y^6}{y^5}=\frac{y^6-1}{y^5}.$

    ✓Final answer

    The correct option is (B), $\dfrac{y^6-1}{y^5}$.