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

Showing the 12 most recent of 27 on this concept.

1. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.Let $y = (\tan x)^{\sin x}$ for $0 < x < \dfrac{\pi}{2}$. If $\dfrac{dy}{dx} = (\tan x)^{\sin x}\left((\cos x)\log(\tan x) + g(x)\right)$, then $g(x) =$ (A) $\sin x \sec^2 x$ (B) $\sec x \operatorname{cosec} x$ (C) $\sec x$ (D) $\operatorname{cosec} x$ (E) $\sin x \tan x$

   ›Reveal solutionSolution

   $g(x)=\sec x$.

   Concept and Intuition

   For a variable base raised to a variable power, take logs first. Differentiating produces a term from the exponent and a term from the base.

   Step-by-Step Solution

   1. $\log y=\sin x\,\log(\tan x)$.
   2. $\dfrac{y'}{y}=\cos x\,\log(\tan x)+\sin x\cdot\dfrac{\sec^2 x}{\tan x}$.
   3. Simplify the second term: $\sin x\cdot\dfrac{\sec^2 x}{\tan x}=\sin x\cdot\dfrac{1}{\cos^2 x}\cdot\dfrac{\cos x}{\sin x}=\dfrac{1}{\cos x}=\sec x$. …

   🔒 Start your 14-day free trial to see the full solution →
2. KEAM 2023Set eng-2023-P2-B24 marksMCQ
   Q.If $y = x^{e^{x}} + x^{e}$ for $x > 0$, then $\dfrac{dy}{dx}$ is equal to (A) $x^{e^{x}}\left[\dfrac{1}{x} + \ln x\right] + e^{x}$ (B) $x^{e^{x}} e^{x}\left[\dfrac{1}{x} + \ln x\right] + e x^{e-1}$ (C) $e^{x} \cdot x^{e^{x} - 1} + e x^{e}$ (D) $x^{e^{x}} e^{-x}\left[\dfrac{1}{x} - \ln x\right] + e x^{e-1}$ (E) $x^{e^{x}} e^{x}\left[\dfrac{1}{x} - \ln x\right] + e x^{e-1}$

   ›Reveal solutionSolution

   $\dfrac{dy}{dx}=x^{e^x}e^x\left[\dfrac1x+\log x\right]+e\,x^{e-1}$.

   Concept and Intuition

   Differentiate the two terms separately: $x^{e^x}$ via logarithmic differentiation, and $x^e$ via the power rule.

   Step-by-Step Solution

   1. Let $u=x^{e^x}$. Then $\log u=e^x\log x$, so $\dfrac{u'}{u}=e^x\log x+e^x\cdot\dfrac1x=e^x\left(\log x+\dfrac1x\right)$.
   2. Thus $u'=x^{e^x}e^x\left(\dfrac1x+\log x\right)$.
   3. For the second term $\dfrac{d}{dx}x^e=e\,x^{e-1}$. Add them.

   Common Mistakes …

   🔒 Start your 14-day free trial to see the full solution →
3. 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

   $f'(x)=\sec^{2}x.$ …

   🔒 Start your 14-day free trial to see the full solution →
4. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.If $x = \sqrt{10^{\cos^{-1}\theta}}$ and $y = \sqrt{10^{\sin^{-1}\theta}}$, then $\frac{dy}{dx}$ is equal to (A) $xy$ (B) $\frac{x}{y}$ (C) $\frac{y}{x}$ (D) $\frac{-x}{y}$ (E) $\frac{-y}{x}$

   ›Reveal solutionSolution

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

   Concept and Intuition

   Both $x$ and $y$ depend on $\theta$; use logarithmic differentiation and the fact that $\frac{d}{d\theta}\cos^{-1}\theta$ and $\frac{d}{d\theta}\sin^{-1}\theta$ are negatives of each other.

   Step-by-Step Solution

   1. $x=10^{\frac{1}{2}\cos^{-1}\theta}$, so $\log x=\frac{1}{2}\log 10\,\cos^{-1}\theta$.
   2. Differentiate: $\frac{1}{x}\frac{dx}{d\theta}=\frac{1}{2}\log 10\cdot\frac{-1}{\sqrt{1-\theta^2}}$.
   3. Similarly $\frac{1}{y}\frac{dy}{d\theta}=\frac{1}{2}\log 10\cdot\frac{+1}{\sqrt{1-\theta^2}}$. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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: $\dfrac{dy}{dx}=\dfrac{1}{2}\left[\dfrac{1}{x-1}-\dfrac{1}{x+2}\right]$. …

   🔒 Start your 14-day free trial to see the full solution →
6. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.Let $f(x)=x\sin(x^{4})$. Then $f'(x)$ at $x=\sqrt[4]{\pi}$ is equal to (A) $4\pi+1$ (B) $4\pi$ (C) $-4\pi$ (D) $4\pi-1$ (E) $4\pi+4$

   ›Reveal solutionSolution

   Product + chain rule, evaluate at $x^4=\pi$.

   $f'(x)=\sin(x^4)+x\cdot\cos(x^4)\cdot4x^3=\sin(x^4)+4x^4\cos(x^4).$ …

   🔒 Start your 14-day free trial to see the full solution →
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. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 $\log y=\sin\dfrac{\pi x}{2}\,\log x$.

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

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $\dfrac{\log_{10}e}{x}$).

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

   🔒 Start your 14-day free trial to see the full solution →
10. KEAM 2024Set eng-2024-06084 marksMCQ
    Q.If $x=r\cos\theta$, $y=r\sin\theta$, then $\dfrac{dy}{dx}$ at $\theta=\dfrac{\pi}{4}$, where $r$ is a constant and $\theta$ is a parameter, is equal to (A) $0$ (B) $1$ (C) $-1$ (D) $\sqrt{2}$ (E) $\dfrac{1}{\sqrt{2}}$

    ›Reveal solutionSolution

    With $r$ constant, $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=-\cot\theta$, which is $-1$ at $\theta=\frac{\pi}{4}$.

    Since $r$ is constant and $\theta$ is the parameter,

    $\frac{dx}{d\theta}=-r\sin\theta,\qquad \frac{dy}{d\theta}=r\cos\theta.$

    Therefore …

    🔒 Start your 14-day free trial to see the full solution →
11. 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, …

    🔒 Start your 14-day free trial to see the full solution →
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: …

    🔒 Start your 14-day free trial to see the full solution →