Q.Find $\frac{dy}{dx}$ in the following: $x \cdot \cos x$

Showing the 12 most recent of 19 on this concept.

1. 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

   $\frac{d}{dx}x^{\log 5}=\log 5\cdot x^{\log 5-1}=\log 5\cdot x^{\log 5-\log e}=\log 5\cdot x^{\log(5/e)}.$

   ✓Final answer

   (C) $\log 5\cdot x^{\log\left(\frac{5}{e}\right)}$

   ANSWER: (C)

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

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

   Writing in radians, $\sin(3x^\circ)=\sin\!\big(\tfrac{\pi x}{60}\big)$, so

   $\frac{d}{dx}\sin\!\Big(\frac{\pi x}{60}\Big)=\frac{\pi}{60}\cos(3x^\circ).$

   ✓Final answer

   (C) $\dfrac{\pi}{60}\cos(3x^\circ)$

   ANSWER: (C)

3. 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

   $[J]=\dfrac{[I]}{[A]}=\dfrac{\text{A}}{\text{m}^{2}}=A\,m^{-2}$. Among the options, (A) $A\,m^{-2}$ is correct (options (B) and (C) are $A\,m^{-1}$, and (D) is $A\,m^{2}$).

   [!ANSWER]

   (A) $Am^{-2}$

   Note

   This solution was worked out by our team and independently cross-checked by a second solve. The official answer key on record for this question could not be confirmed, so please cross-verify with the official paper where possible.

4. 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

   $\frac{dy}{dx}=\frac{a(1+\cos\theta)}{a\sin\theta}=\frac{2\cos^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}=\cot\frac{\theta}{2}.$

   ✓Final answer

   (B) $\cot\frac{\theta}{2}$

   ANSWER: (B)

5. GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
   Q.If $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$, then $\frac{dy}{dx} =$ ______.

   (a) $y$

   (b) $y - 1$

   (c) $0$

   (d) does not exist

   ›Reveal solutionSolution

   This series is exactly the Maclaurin expansion of $e^x$.

   $y=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots=e^x$.

   $\frac{dy}{dx}=e^x=y$.

   ✓Final answer

   The correct option is (a) $y$.

6. 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 $\approx 4\,m_p$).

   Since $\lambda \propto 1/\sqrt{m}$ at fixed KE, the smallest mass gives the largest $\lambda$. The electron, having by far the least mass, has the longest de Broglie wavelength.

   [!ANSWER]

   (C) electron

7. 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

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

   So the ratio of half-life to average life equals $\ln 2$.

   [!ANSWER]

   (C) $\ln(2)$

8. GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 mark
   Q.If $f(x) = 9x^2 - 8x + 6$, find $f'(x)$.

   ›Reveal solutionSolution

   Differentiate term by term using the power rule $\dfrac{d}{dx}(x^n)=nx^{n-1}$.

   $f(x) = 9x^2-8x+6$. Differentiating term by term:

   $f'(x) = 9(2x) - 8(1) + 0 = 18x - 8.$

   ✓Final answer

   $f'(x) = 18x-8$.

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

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

   $\log y=\tfrac{\ln10}{2}\cos^{-1}t\Rightarrow \tfrac1y\tfrac{dy}{dt}=\tfrac{\ln10}{2}\cdot\tfrac{-1}{\sqrt{1-t^2}}$.

   Dividing, $\dfrac{dy}{dx}=\dfrac{y\cdot(-1)}{x\cdot(1)}=-\dfrac{y}{x}$.

   ✓Final answer

   (D) $-\frac{y}{x}$

   ANSWER: (D)

10. GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
    Q.The slope of the tangent to the curve $x = t^2 + 3t - 8$, $y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is ___.

    (a) $-\frac{7}{6}$

    (b) $\frac{7}{6}$

    (c) $\frac{6}{7}$

    (d) $-\frac{6}{7}$

    ›Reveal solutionSolution

    For a parametric curve the slope is $\tfrac{dy/dt}{dx/dt}$; find $t$ at the point, then evaluate.

    From $x = t^2+3t-8 = 2 \Rightarrow t^2+3t-10=0 \Rightarrow t=2$ or $t=-5$.

    From $y = 2t^2-2t-5 = -1 \Rightarrow t^2-t-2=0 \Rightarrow t=2$ or $t=-1$.

    Common value: $t=2$.

    $\dfrac{dx}{dt}=2t+3,\quad \dfrac{dy}{dt}=4t-2 \Rightarrow \dfrac{dy}{dx}=\dfrac{4t-2}{2t+3}.$

    At $t=2$: $\dfrac{8-2}{4+3}=\dfrac{6}{7}.$

    ✓Final answer

    (c) $\tfrac{6}{7}$.

11. 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

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

    $\frac{d}{dx}\big(4x^3+3x^2+3x+4\big)=12x^2+6x+3.$

    ✓Final answer

    (D) $12x^2 + 6x + 3$

    ANSWER: (D)

12. 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:

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

    $\frac{d}{dx}(-6\log x)=-\frac{6}{x}.$

    ✓Final answer

    (A) $-\dfrac{6}{x}$

    ANSWER: (A)