Q.Find $\frac{dy}{dx}$ in the following: $\frac{e^x}{\sin x}$

Showing the 12 most recent of 27 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The point $P(x, y)$, where $y = 4\log_e(2)$, lies on the curve with equation $y = \log_e(x^3 + 24)$. Then the value of $\dfrac{dy}{dx}$ at the point $P$ is (A) $\dfrac{-3}{8}$ (B) $\dfrac{3}{8}$ (C) $\dfrac{-3}{4}$ (D) $\dfrac{3}{4}$ (E) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Find $x$ from $y=4\log_e2$, then evaluate $\dfrac{dy}{dx}=\dfrac{3x^2}{x^3+24}$.

   Since $y=4\log_e2=\log_e16$ and $y=\log_e(x^3+24)$, we get $x^3+24=16\Rightarrow x^3=-8\Rightarrow x=-2$.

   Differentiating $y=\log_e(x^3+24)$: $\dfrac{dy}{dx}=\dfrac{3x^2}{x^3+24}$.

   At $x=-2$: $\dfrac{3(4)}{-8+24}=\dfrac{12}{16}=\dfrac34$.

   ✓Final answer

   The correct option is (D).

2. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The point P lies on the curve with equation $y = x\sqrt{2\log_e(x)},\ x > 0$. If $\dfrac{dy}{dx} = 2$ at $x = k$, then the value of $k$ is equal to (A) $-\sqrt{e}$ (B) $e$ (C) $\sqrt{e}$ (D) $2e$ (E) $\dfrac{e}{2}$

   ›Reveal solutionSolution

   Differentiating $y=x\sqrt{2\log x}$ gives $\sqrt{2\log x}+\dfrac{1}{\sqrt{2\log x}}$; setting it to $2$ gives $2\log x=1$, i.e. $x=\sqrt e$.

   Write $y=x\,u$ with $u=(2\log x)^{1/2}$. Then

   $u'=\tfrac12(2\log x)^{-1/2}\cdot\tfrac{2}{x}=\dfrac{1}{x\sqrt{2\log x}},$

   so $\dfrac{dy}{dx}=u+x\,u'=\sqrt{2\log x}+\dfrac{1}{\sqrt{2\log x}}$.

   Let $w=\sqrt{2\log x}$. Setting $w+\dfrac1w=2$ gives $w^2-2w+1=0$, i.e. $(w-1)^2=0$, so $w=1$. Then $2\log x=1$, $\log x=\tfrac12$, and $x=k=\sqrt{e}$.

   ✓Final answer

   The correct option is (C).

3. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.Let $y = x^{\sin\frac{\pi}{2}x}$. Then at $x = 1$, $\frac{dy}{dx}$ is equal to (A) $-1$ (B) $0$ (C) $-2$ (D) $2$ (E) $1$

   ›Reveal solutionSolution

   Take logs, differentiate, and evaluate at $x=1$ where $\log 1=0$ kills one term.

   $y=x^{\sin\frac{\pi}{2}x}$, so $\log y=\sin\dfrac{\pi x}{2}\,\log x$.

   $\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{\pi}{2}\cos\dfrac{\pi x}{2}\,\log x+\sin\dfrac{\pi x}{2}\cdot\dfrac{1}{x}$.

   At $x=1$: $\cos\dfrac{\pi}{2}=0$ (first term vanishes) and $\sin\dfrac{\pi}{2}=1$, so $\dfrac{y'}{y}=1$. Since $y(1)=1^1=1$, $\dfrac{dy}{dx}=1$.

   ✓Final answer

   The correct option is (E).

4. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.If $y = \log\left(\sqrt{\dfrac{x-1}{x+2}}\right)$ then $\dfrac{dy}{dx} =$ (A) $\dfrac{1}{2(x-1)(x+2)}$ (B) $\dfrac{3}{2(x-1)(x+2)}$ (C) $\dfrac{3}{(x-1)(x+2)}$ (D) $\dfrac{1}{(x-1)(x+2)}$ (E) $\dfrac{1}{3(x-1)(x+2)}$

   ›Reveal solutionSolution

   Use log rules to split, differentiate, and combine: $\dfrac{dy}{dx}=\dfrac{3}{2(x-1)(x+2)}$.

   $y=\log\sqrt{\dfrac{x-1}{x+2}}=\dfrac{1}{2}\big[\log(x-1)-\log(x+2)\big]$.

   Differentiate: $\dfrac{dy}{dx}=\dfrac{1}{2}\left[\dfrac{1}{x-1}-\dfrac{1}{x+2}\right]$.

   Combine: $\dfrac{1}{x-1}-\dfrac{1}{x+2}=\dfrac{(x+2)-(x-1)}{(x-1)(x+2)}=\dfrac{3}{(x-1)(x+2)}$.

   So $\dfrac{dy}{dx}=\dfrac{3}{2(x-1)(x+2)}$.

   ✓Final answer

   The correct option is (B).

5. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.If $y = \log_{10} x + \log_e x$, then $\dfrac{dy}{dx}$ is equal to (A) $\dfrac{1 - \log_{10} e}{x}$ (B) $\dfrac{1 + \log_e 10}{x}$ (C) $x + \log_{10} e$ (D) $x + \log_e 10$ (E) $\dfrac{1}{x}\left[\dfrac{1}{\log_e 10} + 1\right]$

   ›Reveal solutionSolution

   Differentiate each log term; $\dfrac{dy}{dx}=\dfrac{1}{x}\left[\dfrac{1}{\log_e 10}+1\right]$.

   $\log_{10}x=\dfrac{\log_e x}{\log_e 10}$, so $\dfrac{d}{dx}\log_{10}x=\dfrac{1}{x\log_e 10}$ (equivalently $\dfrac{\log_{10}e}{x}$).

   $\dfrac{d}{dx}\log_e x=\dfrac{1}{x}$.

   Adding: $\dfrac{dy}{dx}=\dfrac{1}{x\log_e 10}+\dfrac{1}{x}=\dfrac{1}{x}\left[\dfrac{1}{\log_e 10}+1\right]$.

   ✓Final answer

   The correct option is (E).

6. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.Let $f(x) = |\sin 3x| - |\cos 3x|$, where $\dfrac{\pi}{6} \le x \le \dfrac{\pi}{3}$. Then the value of $f'\left(\dfrac{\pi}{4}\right)$ is equal to (A) $-3\sqrt{2}$ (B) $3\sqrt{2}$ (C) $\dfrac{-3}{\sqrt{2}}$ (D) $\dfrac{3}{\sqrt{2}}$ (E) $0$

   ›Reveal solutionSolution

   Near x=pi/4, f simplifies to sin3x+cos3x, giving f'(pi/4) = -3*sqrt(2).

   Concept and Intuition

   To differentiate an absolute value, first resolve the signs of the inner functions on the relevant interval, then differentiate the resulting smooth expression. Here 3x = 3pi/4 lies in the second quadrant.

   Step-by-Step Solution

   1. At x = pi/4, 3x = 3pi/4, where sin3x > 0 and cos3x < 0.
   2. So |sin3x| = sin3x and |cos3x| = -cos3x, giving f(x) = sin3x - (-cos3x) = sin3x + cos3x.
   3. Differentiate: f'(x) = 3cos3x - 3sin3x.
   4. At 3x = 3pi/4: cos = -sqrt(2)/2, sin = sqrt(2)/2, so f' = 3(-sqrt(2)/2) - 3(sqrt(2)/2) = -3*sqrt(2).

   Common Mistakes

   • Dropping the sign flip on |cos3x| when cos3x is negative.

   ✓Final answer

   The correct option is (A) — -3*sqrt(2).

   ANSWER: A

7. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.Let $f(x) = (\cos^2 x)(a + \cos x)$. If $f'\left(\dfrac{\pi}{3}\right) = 0$ then the value of $a$ is equal to (A) $\dfrac{\sqrt{3}}{2}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{-3}{4}$ (D) $\dfrac{-3}{2}$ (E) $-1$

   ›Reveal solutionSolution

   Setting f'(pi/3)=0 for f=cos^2 x (a+cos x) gives a = -3/4.

   Concept and Intuition

   Differentiate the product cos^2 x times (a+cos x), evaluate at pi/3 using cos(pi/3)=1/2, sin(pi/3)=sqrt(3)/2, and solve for a.

   Step-by-Step Solution

   1. f'(x) = 2cos x(-sin x)(a+cos x) + cos^2 x(-sin x) = -2 cos x sin x (a+cos x) - cos^2 x sin x.
   2. At x = pi/3: cos = 1/2, sin = sqrt(3)/2.
   3. f' = -2(1/2)(sqrt3/2)(a+1/2) - (1/4)(sqrt3/2) = -(sqrt3/2)(a+1/2) - sqrt3/8.
   4. Set = 0 and divide by sqrt3: -(1/2)(a+1/2) - 1/8 = 0, i.e. -a/2 - 1/4 - 1/8 = 0, so a/2 = -3/8, a = -3/4.

   Common Mistakes

   • Missing the second product-rule term cos^2 x * (-sin x).

   ✓Final answer

   The correct option is (C) — -3/4.

   ANSWER: C

8. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.If $y=\sec\left(\tan^{-1}x\right)$, then $\dfrac{dy}{dx}$ at $x=\sqrt{3}$ is equal to (A) $\dfrac{2\sqrt{3}}{3}$ (B) $\dfrac{1}{2}$ (C) $2$ (D) $\dfrac{\sqrt{3}}{2}$ (E) $2\sqrt{3}$

   ›Reveal solutionSolution

   Simplify $\sec(\tan^{-1}x)=\sqrt{1+x^{2}}$, differentiate to $\dfrac{x}{\sqrt{1+x^{2}}}$, then substitute $x=\sqrt{3}$.

   Let $\theta=\tan^{-1}x$ so $\tan\theta=x$. Then $\sec\theta=\sqrt{1+\tan^{2}\theta}=\sqrt{1+x^{2}}$, hence

   $y=\sqrt{1+x^{2}}.$

   Differentiating,

   $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{1+x^{2}}}\cdot 2x=\dfrac{x}{\sqrt{1+x^{2}}}.$

   At $x=\sqrt{3}$: $\dfrac{\sqrt{3}}{\sqrt{1+3}}=\dfrac{\sqrt{3}}{2}.$

   ✓Final answer

   The correct option is (D).

9. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.If $f(x)=\dfrac{\sqrt{2}\sin x}{\sqrt{1+\cos 2x}}$, then $f'\left(\dfrac{\pi}{6}\right)=$ (A) $\dfrac{1}{4}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{4}{3}$ (D) $\dfrac{1}{2}$ (E) $\dfrac{3}{4}$

   ›Reveal solutionSolution

   Use $1+\cos 2x=2\cos^{2}x$ to simplify $f$ to $\tan x$ near $x=\pi/6$, then $f'=\sec^{2}x$.

   Since $1+\cos 2x=2\cos^{2}x$,

   $f(x)=\dfrac{\sqrt{2}\sin x}{\sqrt{2\cos^{2}x}}=\dfrac{\sqrt{2}\sin x}{\sqrt{2}\,|\cos x|}=\dfrac{\sin x}{|\cos x|}.$

   Near $x=\pi/6$, $\cos x>0$, so $f(x)=\tan x$ and

   $f'(x)=\sec^{2}x.$

   At $x=\pi/6$: $\sec^{2}\dfrac{\pi}{6}=\dfrac{1}{\cos^{2}(\pi/6)}=\dfrac{1}{3/4}=\dfrac{4}{3}.$

   ✓Final answer

   The correct option is (C).

10. KEAM 2025Set eng-2025-04284 marksMCQ
    Q.If $y = \sin^{-1}\left(2x\sqrt{1 - x^2}\right)$, then $\frac{dy}{dx}$ at $x=0$ is (A) $0$ (B) $1$ (C) $2$ (D) $\frac{\sqrt{3}}{2}$ (E) $-1$

    ›Reveal solutionSolution

    Substituting $x=\sin\theta$ gives $y=2\sin^{-1}x$, so $y'=\dfrac{2}{\sqrt{1-x^2}}=2$ at $x=0$.

    Let $x=\sin\theta$. Then $2x\sqrt{1-x^2}=2\sin\theta\cos\theta=\sin 2\theta$, so near $x=0$,

    $y=\sin^{-1}(\sin2\theta)=2\theta=2\sin^{-1}x.$

    Differentiating,

    $\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}.$

    At $x=0$,

    $\frac{dy}{dx}=\frac{2}{\sqrt{1}}=2.$

    ✓Final answer

    The correct option is (C).

11. KEAM 2025Set eng-2025-04294 marksMCQ
    Q.If $\log_2 y = x$, then $\dfrac{dy}{dx}$ is equal to (A) $2^x \log_e 2$ (B) $2^x$ (C) $x^2$ (D) $2x$ (E) $\dfrac{2^x}{\log_e y}$

    ›Reveal solutionSolution

    $\log_2 y=x$ means $y=2^x$; differentiating $2^x$ gives $2^x\log_e 2$.

    From $\log_2 y=x$ we get $y=2^x$.

    Differentiating an exponential $a^x$ gives $a^x\ln a$, so

    $\frac{dy}{dx}=2^x\log_e 2.$

    ✓Final answer

    The correct option is (A).

12. KEAM 2025Set eng-2025-04294 marksMCQ
    Q.The derivative of $y = (x - 1)(2x - 1)(3 - x)(4 - x)$ at $x = \dfrac{1}{2}$ is equal to (A) $35$ (B) $\dfrac{-35}{4}$ (C) $\dfrac{-35}{2}$ (D) $\dfrac{35}{4}$ (E) $\dfrac{35}{2}$

    ›Reveal solutionSolution

    Since $(2x-1)=0$ at $x=\frac12$, every product-rule term keeping that factor vanishes; only differentiating $(2x-1)$ survives, giving $y'=-\frac{35}{4}$.

    Write $y=(x-1)(2x-1)(3-x)(4-x)$. By the product rule, $y'$ is a sum of four terms, each differentiating one factor and keeping the others. At $x=\frac12$ the factor $(2x-1)=0$, so every term still containing $(2x-1)$ is zero. Only the term differentiating $(2x-1)$ (derivative $=2$) remains:

    $y'\Big|_{1/2}=(x-1)\cdot2\cdot(3-x)(4-x)\Big|_{1/2}.$

    Compute:

    $=\left(-\tfrac12\right)(2)\left(\tfrac52\right)\left(\tfrac72\right)=(-1)\cdot\frac{35}{4}=-\frac{35}{4}.$

    ✓Final answer

    The correct option is (B).