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

Showing the 12 most recent of 19 on this concept.

1. GUJCET 2025Set 031 markMCQ
   Q.$\frac{d}{dx}\left[3\sin(60° - x°) - 4\cos^3(30° + x°)\right] =$ _____. (A) $-\frac{\pi}{60}\sin(3x°)$ (B) $\frac{\pi}{60}\sin(3x°)$ (C) $\frac{\pi}{60}\cos(3x°)$ (D) $-\frac{\pi}{60}\cos(3x°)$

   ›Reveal solutionSolution

   With $u=60^\circ-x^\circ$, $\cos(30^\circ+x^\circ)=\sin u$, so the expression is $\sin3u=\sin(3x^\circ)$.

   Let $u=60^\circ-x^\circ$. Since $\cos(30^\circ+x^\circ)=\sin(60^\circ-x^\circ)=\sin u$:

   $3\sin u-4\sin^3 u=\sin3u=\sin(180^\circ-3x^\circ)=\sin(3x^\circ).$ …

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2. GUJCET 2024Set 131 markMCQ
   Q.If $x = a(1 - \cos\theta), y = a(\theta + \sin\theta)$ then $\frac{dy}{dx} =$ __________. (A) $-\tan\frac{\theta}{2}$ (B) $\cot\frac{\theta}{2}$ (C) $-\cot\frac{\theta}{2}$ (D) $\tan\frac{\theta}{2}$

   ›Reveal solutionSolution

   Using parametric differentiation, $\dfrac{dy}{dx}=\dfrac{1+\cos\theta}{\sin\theta}=\cot\dfrac{\theta}{2}$.

   Concept. $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$, then simplify with half-angle identities.

   Steps. $\dfrac{dx}{d\theta}=a\sin\theta$, $\dfrac{dy}{d\theta}=a(1+\cos\theta)$, so …

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3. GUJCET 2022Set 081 markMCQ
   Q.If $x=\sqrt{10^{\sin^{-1}t}}$, $y=\sqrt{10^{\cos^{-1}t}}$ then $\frac{dy}{dx}=$ ______. (A) $-\frac{x}{y}$ (B) $\frac{y}{x}$ (C) $0$ (D) $-\frac{y}{x}$

   ›Reveal solutionSolution

   Since $\sin^{-1}t+\cos^{-1}t=\tfrac\pi2$, the two exponents move oppositely, giving $dy/dx=-y/x$.

   Concept. $x=10^{(\sin^{-1}t)/2}$, $y=10^{(\cos^{-1}t)/2}$.

   $\log x=\tfrac{\ln10}{2}\sin^{-1}t\Rightarrow \tfrac1x\tfrac{dx}{dt}=\tfrac{\ln10}{2}\cdot\tfrac1{\sqrt{1-t^2}}$. …

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4. GUJCET 2025Set 031 markMCQ
   Q.$\frac{d}{dx}\left(5^{\log x}\right) =$ _____ (A) $\log 5 \cdot x^{\log(5e)}$ (B) $\log_x 5 \cdot 5^{\log x}$ (C) $\log 5 \cdot x^{\log\left(\frac{5}{e}\right)}$ (D) $\log 5 \cdot 5^{\log x}$

   ›Reveal solutionSolution

   Rewrite $5^{\log x}=x^{\log 5}$ and differentiate the power.

   $5^{\log x}=e^{\log x\,\log 5}=x^{\log 5}$. Then …

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5. GUJCET 2021Set 151 markMCQ
   Q.If $f(x) = 4x^3 + 3x^2 + 3x + 4$; $x \ne 0$, then $\frac{d}{dx}\left(x^3 \cdot f\left(\frac{1}{x}\right)\right) = \underline{\hspace{2cm}}$. (A) $24x^5 + 15x^4 + 12x^3 + 12x^2$ (B) $\frac{12}{x^2} + \frac{6}{x} + 3$ (C) $\frac{x^2}{12} + \frac{x}{6} + \frac{1}{3}$ (D) $12x^2 + 6x + 3$

   ›Reveal solutionSolution

   Substitute $1/x$, multiply by $x^3$, then differentiate.

   Concept: $f(1/x)=\dfrac{4}{x^3}+\dfrac{3}{x^2}+\dfrac{3}{x}+4$, so

   $x^3 f\!\left(\tfrac1x\right)=4+3x+3x^2+4x^3.$ …

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6. GUJCET 2020Set 071 markMCQ
   Q.If $y = \sin^{-1}\left(\frac{2^{x+1}}{1 + 4^x}\right)$ and $\frac{dy}{dx} = \frac{2^{x+1}\log 2}{f(x)}$ then $f(0) =$ ________. (A) $0$ (B) $-2$ (C) $2$ (D) $2\log 2$

   ›Reveal solutionSolution

   $\sin^{-1}\!\frac{2t}{1+t^2}=2\tan^{-1}t$ with $t=2^x$ gives $f(x)=1+4^x$, so $f(0)=2$.

   Concept — inverse-trig substitution. Let $t=2^x$. Then $\dfrac{2^{x+1}}{1+4^x}=\dfrac{2t}{1+t^2}$ and $\sin^{-1}\dfrac{2t}{1+t^2}=2\tan^{-1}t$. …

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7. GUJCET 2021Set 151 markMCQ
   Q.$\frac{d}{dx}\left[\log\left(\frac{1}{x}\right) + \log\left(\frac{1}{x^2}\right) + \log\left(\frac{1}{x^3}\right)\right] = \underline{\hspace{2cm}}$; $x > 1$. (A) $-\frac{6}{x}$ (B) $6x$ (C) $\frac{6}{x}$ (D) $-6x$

   ›Reveal solutionSolution

   $\log(1/x^k)=-k\log x$; sum the coefficients.

   Concept:

   $\log\tfrac1x+\log\tfrac1{x^2}+\log\tfrac1{x^3}=-(1+2+3)\log x=-6\log x.$ …

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8. GUJCET 2021Set 151 markMCQ
   Q.The slope of normal to the curve $y = 2x^2 + 3\sin x$ at $x = 0$ is $\underline{\hspace{2cm}}$. (A) $3$ (B) $-3$ (C) $\frac{1}{3}$ (D) $-\frac{1}{3}$

   ›Reveal solutionSolution

   $y'=4x+3\cos x$ gives tangent slope $3$ at $x=0$; the normal slope is $-\dfrac{1}{3}$.

   Concept: $\dfrac{dy}{dx}=4x+3\cos x$. At $x=0$: slope $=0+3=3$. …

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9. GUJCET 2023Set 091 markMCQ
   Q.$\left\{\frac{d}{dx}\left(x^{x} + x^{x+1} + x^{x+2}\right)\right\}_{x=e} =$ ______. (A) $e^{e}\left(1 + e^{2} + 2e\right)$ (B) $e^{e}\left(3e^{2} + 2e + 2\right)$ (C) $e^{e}\left(2e^{2} + 4e + 3\right)$ (D) $e^{e}\left(1 + 4e + 2e^{2}\right)$

   ›Reveal solutionSolution

   [!TLDR]

   With equal KE, the lightest particle (electron) has the longest de Broglie wavelength.

   Concept

   The de Broglie wavelength in terms of kinetic energy is $\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mE}}$. For the same $E$, $\lambda \propto \dfrac{1}{\sqrt{m}}$.

   Solution

   Masses: $m_{e} \ll m_{p} < m_{\alpha}$ (proton $\approx 1836\,m_e$; $\alpha$-particle $\approx 4\,m_p$). …

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10. GUJCET 2024Set 131 markMCQ
    Q.$\frac{d}{dx}\left(e^{x \log x} + e^3\right) =$ __________. (A) $x^x(1 + \log x) + e^3$ (B) $(1 + \log x)$ (C) $x^x \log x$ (D) $x^x(1 + \log x)$

    ›Reveal solutionSolution

    [!TLDR]

    $J=I/A$ has units of ampere per square metre, $A\,m^{-2}$ — option (A).

    Concept

    Current density is the electric current per unit cross-sectional area through which it flows: $J=\dfrac{I}{A}$. Current is in amperes (A) and area in square metres (m$^2$).

    Solution …

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11. GUJCET 2023Set 091 markMCQ
    Q.An approximate value of $(81.5)^{1/4}$ is : (A) $3.0436$ (B) $3.0033$ (C) $3.0046$ (D) $3.0465$

    ›Reveal solutionSolution

    [!TLDR]

    $T_{1/2}/\tau = \ln 2 \approx 0.693$.

    Concept

    For radioactive decay with constant $\lambda$: half-life $T_{1/2} = \dfrac{\ln 2}{\lambda}$ and mean (average) life $\tau = \dfrac{1}{\lambda}$.

    Solution …

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12. GUJCET 2019Set 171 markMCQ
    Q.The approximate value of $5^{2.01}$ is $\underline{\hspace{2cm}}$, where, $(\log_e 5 = 1.6095)$ (A) $25.2525$ (B) $25.5025$ (C) $25.4125$ (D) $25.4024$

    ›Reveal solutionSolution

    Using $f(x)=5^x$, $5^{2.01}\approx f(2)+0.01f'(2)=25+0.01(25\ln5)=25.4024$.

    Concept: $f(x)=5^x\Rightarrow f'(x)=5^x\ln5$. At $x=2$: $f(2)=25$, $f'(2)=25\times1.6095=40.2375$. …

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