Q.Find $\frac{dy}{dx}$ in the following: $e^x \sin 5x$

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2024Set 2024-E1 markMCQ
   Q.

   [!FORMULA] $\text { If } y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^2}}{13}\right) \text { then } \frac{d y}{d x} \text { equals }$

   (A) $\frac{-2 x}{\sqrt{1-x^2}}$ (B) $\frac{-1}{\sqrt{1+x^2}}$ (C) $\frac{1}{\sqrt{1-x^2}}$ (D) $\frac{2 x}{\sqrt{1-x^2}}$

   ›Reveal solutionSolution

   Substituting $x=\sin\theta$ turns the argument into $\sin(\theta+\phi)$, so $y=\sin^{-1}x+\phi$ and $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$. The correct option is (C).

   Concept

   When an inverse-sine argument has the form $\frac{a x+b\sqrt{1-x^2}}{c}$ with $a^2+b^2=c^2$, the substitution $x=\sin\theta$ collapses $a\sin\theta+b\cos\theta$ into $c\sin(\theta+\phi)$, so the inverse sine unwinds to a simple sum of an angle and a constant — making differentiation immediate.

   Solution

   1. Substitute. $x=\sin\theta$, $\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$, so $\sqrt{1-x^2}=\cos\theta$ and

   $y=\sin^{-1}\!\left(\frac{5\sin\theta+12\cos\theta}{13}\right).$

   2. Single sine. With $\cos\phi=\tfrac{5}{13},\ \sin\phi=\tfrac{12}{13}$, $5\sin\theta+12\cos\theta=13\sin(\theta+\phi),\qquad y=\sin^{-1}\big(\sin(\theta+\phi)\big).$ …

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2. KCET 2021Set A-11 markMCQ
   Q.If $y = (\cos x^2)^2$, then $\frac{dy}{dx}$ is equal to (A) $-4x \sin 2x^2$ (B) $-x \sin x^2$ (C) $-2x \sin 2x^2$ (D) $-x \cos 2x^2$

   ›Reveal solutionSolution

   Use the chain rule twice: differentiate the outer square, then the cosine, then the inner $x^2$. The derivative is $-4x \sin(x^2) \cos(x^2)$, which simplifies to $-2x \sin(2x^2)$.

   The function $y = (\cos x^2)^2$ is a composition of three layers: an outer square, a middle cosine, and an innermost $x^2$. Whenever you see a function of a function of a function, the chain rule is your tool — you differentiate from the outside in, multiplying each derivative along the way.

   A common mistake is to forget that $\cos x^2$ means $\cos(x^2)$, not $(\cos x)^2$. Here the parentheses make it clear: $(\cos x^2)^2$ means "square the cosine of $x^2$". So the outermost operation is squaring, then cosine, then $x^2$.

   Let’s work through it step by step.

   1. Identify the layers.

      Write $y = u^2$ where $u = \cos v$ and $v = x^2$.

      Then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}$.

   2. Differentiate each layer.

      • $\frac{dy}{du} = 2u = 2\cos v = 2\cos(x^2)$
      • $\frac{du}{dv} = -\sin v = -\sin(x^2)$
      • $\frac{dv}{dx} = 2x$

   3. Multiply them together.

   $\frac{dy}{dx} = 2\cos(x^2) \cdot \bigl(-\sin(x^2)\bigr) \cdot 2x$

   $= -4x \,\cos(x^2)\,\sin(x^2)$

   4. Simplify using a trig identity. Recall the double-angle identity: $\sin 2\theta = 2\sin\theta\cos\theta$. Here $\theta = x^2$, so $2\sin(x^2)\cos(x^2) = \sin(2x^2)$. Therefore, $-4x \,\cos(x^2)\,\sin(x^2) = -2x \cdot \bigl(2\sin(x^2)\cos(x^2)\bigr) = -2x \sin(2x^2)$ …

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3. KCET 2019Set A-11 markMCQ
   Q.If $3\sqrt{y}\sqrt{x} = \sqrt[6]{(x+y)^5}$, then $\frac{dy}{dx} =$ (A) $\frac{x}{y}$ (B) $x+y$ (C) $x-y$ (D) $\frac{y}{x}$

   ›Reveal solutionSolution

   $\dfrac{dy}{dx} = \dfrac{y}{x}$ — option (D).

   The relation $\sqrt[3]{y}\,\sqrt{x} = \sqrt[6]{(x+y)^5}$ is $x^{1/2}\,y^{1/3} = (x+y)^{1/2 + 1/3}$, the homogeneous form $x^{m}y^{n} = (x+y)^{m+n}$ with $m=\tfrac12,\ n=\tfrac13$.

   Take logarithms: $m\log x + n\log y = (m+n)\log(x+y)$. Differentiate:

   $\dfrac{m}{x} + \dfrac{n}{y}\dfrac{dy}{dx} = \dfrac{(m+n)}{x+y}\left(1 + \dfrac{dy}{dx}\right).$ …

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4. KCET 2024Set A-11 markMCQ
   Q.Let the function satisfy the equation $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$, where $f(0) \neq 0$. If $f(5) = 3$ and $f'(0) = 2$, then $f'(5)$ is (A) $6$ (B) $0$ (C) $3$ (D) $-6$

   ›Reveal solutionSolution

   The functional equation $f(x+y)=f(x)f(y)$ with $f(0)\neq0$ forces $f$ to be an exponential function. Using the given $f(5)=3$ and $f'(0)=2$, we find $f'(5)=6$, so the answer is (A).

   The core idea here is that the equation $f(x+y)=f(x)f(y)$ is the Cauchy exponential functional equation. For functions continuous at even a single point (or differentiable at a point, as we have here), the only non-zero solutions are of the form $f(x)=a^x$ for some positive base $a$. But we don't need to assume continuity — differentiability at $0$ is enough to pin down the derivative everywhere.

   Let’s see why. The equation tells us that the value of $f$ at a sum is the product of its values at the parts. This is the defining property of exponential functions. If we differentiate with respect to $y$ and then set $y=0$, we get a direct relation between $f'(x)$ and $f(x)$.

   1. Differentiate the functional equation with respect to $y$. Treat $x$ as fixed. The left side is $f(x+y)$, whose derivative with respect to $y$ is $f'(x+y)$ (by the chain rule). The right side is $f(x)f(y)$, whose derivative is $f(x)f'(y)$. So:

   $f'(x+y) = f(x)f'(y)$

   This holds for all real $x$ and $y$.

   2. Set $y=0$. Then $f'(x+0) = f(x)f'(0)$. But $f'(0)=2$ is given, so:

   $f'(x) = 2\,f(x)$

   This is a beautiful result: the derivative of $f$ at any point is just twice the function’s value at that point. It tells us $f$ satisfies the differential equation $f' = 2f$, which indeed gives $f(x)=Ce^{2x}$.

   3. Find $f(0)$ using the functional equation. …

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5. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\text { If } y=\sin ^{-1}(\sqrt{\sin x}) \text {, then } \frac{d y}{d x} \text { equals }$

   (A) $\frac{1}{2} \sqrt{1-\operatorname{cosec} x}$ (B) $\frac{1}{2} \sqrt{1-\sin x}$ (C) $\frac{1}{2} \sqrt{1+\operatorname{cosec} x}$ (D) $\frac{1}{2} \sqrt{1+\sin x}$

   ›Reveal solutionSolution

   Chain rule on $y=\sin^{-1}(\sqrt{\sin x})$ simplifies to $\frac{dy}{dx}=\frac12\sqrt{1+\operatorname{cosec}x}$ — option (C).

   Set up the chain rule

   Let $u=\sqrt{\sin x}$, so $y=\sin^{-1}u$ and $u^2=\sin x$.

   $\frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx}.$

   Differentiate $u=(\sin x)^{1/2}$:

   $\frac{du}{dx}=\frac{\cos x}{2\sqrt{\sin x}}.$

   Since $u^2=\sin x$, we have $\sqrt{1-u^2}=\sqrt{1-\sin x}$, so

   $\frac{dy}{dx}=\frac{1}{\sqrt{1-\sin x}}\cdot\frac{\cos x}{2\sqrt{\sin x}}=\frac{\cos x}{2\sqrt{\sin x}\,\sqrt{1-\sin x}}.$

   Simplify with $\cos^2 x=(1-\sin x)(1+\sin x)$

   On the principal domain ($\cos x\ge0$), $\cos x=\sqrt{(1-\sin x)(1+\sin x)}$, hence …

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6. COMEDK 2021Set 2021-B1 markMCQ
   Q.If $f(x)=\log_{x^2}(\log_e x)$, then f'(x) at x = e is (A) infinite (B) 1/2e (C) 0 (D) 1/e

   ›Reveal solutionSolution

   $f'(e)=\dfrac{1}{2e}$.

   Convert to natural logs: $f(x)=\log_{x^2}(\log_e x)=\dfrac{\ln(\ln x)}{\ln(x^2)}=\dfrac{\ln(\ln x)}{2\ln x}$.

   Let $u=\ln(\ln x)$ and $v=2\ln x$, so $f=u/v$ and $f'=\dfrac{u'v-uv'}{v^2}$.

   $u'=\dfrac{1}{\ln x}\cdot\dfrac{1}{x}=\dfrac{1}{x\ln x}$, $\quad v'=\dfrac{2}{x}$. …

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7. COMEDK 2025Set 2025-M1 markMCQ
   Q.If $y=\sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)$ then $\frac{d y}{d x}=$ (A) $\frac{1}{2 \sqrt{1-x}}$ (B) $\frac{1}{2 \sqrt{x}(1+\sqrt{x})}$ (C) $\frac{1}{2 x(1+\sqrt{x})}$ (D) $-\frac{1}{2 \sqrt{x}(1+x)}$

   ›Reveal solutionSolution

   The derivative simplifies by substituting $x = \sec^2\theta - 1$ or using a chain rule with algebraic manipulation; the final result is $-\frac{1}{2\sqrt{x}(1+x)}$, which matches option (D).

   We start with

   $y = \sin^{-1}\left(\frac{1}{\sqrt{x+1}}\right).$

   The argument $\frac{1}{\sqrt{x+1}}$ is always between 0 and 1 for $x \ge 0$, so the inverse sine is well-defined. The key insight: instead of differentiating directly and getting tangled in messy algebra, we can simplify the expression before differentiating by using a trigonometric substitution or by rewriting the argument in a friendlier form.

   ---

   1. Rewrite the argument Let $t = \frac{1}{\sqrt{x+1}}$. Then $t^2 = \frac{1}{x+1}$, so $x+1 = \frac{1}{t^2}$ and $x = \frac{1}{t^2} - 1$. But more directly: notice that

   $\frac{1}{\sqrt{x+1}} = \sqrt{\frac{1}{x+1}}.$

   This suggests setting $x+1 = \sec^2\theta$ (since $\sec\theta \ge 1$ for $\theta \in [0,\pi/2)$). Then

   $\frac{1}{\sqrt{x+1}} = \frac{1}{\sec\theta} = \cos\theta.$

   So

   $y = \sin^{-1}(\cos\theta).$

   2. Simplify the inverse trig expression Recall the identity: $\sin^{-1}(\cos\theta) = \frac{\pi}{2} - \theta$ for $\theta \in [0,\pi]$. Since $\theta = \sec^{-1}(\sqrt{x+1})$ and $\sqrt{x+1} \ge 1$, $\theta$ lies in $[0, \pi/2)$, so the identity holds. Hence

   $y = \frac{\pi}{2} - \theta = \frac{\pi}{2} - \sec^{-1}(\sqrt{x+1}).$

   3. Differentiate The derivative of $\sec^{-1}(u)$ is $\frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{du}{dx}$. Here $u = \sqrt{x+1} > 0$, so the absolute value is unnecessary.

   $\frac{dy}{dx} = 0 - \frac{1}{\sqrt{x+1} \cdot \sqrt{(\sqrt{x+1})^2 - 1}} \cdot \frac{d}{dx}(\sqrt{x+1}).$

   Now $\frac{d}{dx}(\sqrt{x+1}) = \frac{1}{2\sqrt{x+1}}$.

   So

   $$$$

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8. COMEDK 2023Set 2023-E1 markMCQ
   Q.

   [!FORMULA] $\text { If } f(x)=\sin ^{-1}\left(\frac{2^{x+1}}{1+4^x}\right) \text { then } f^{\prime}(0) \text { is equal to }$

   (A) $2 \log 2$ (B) $\frac{2}{3} \log 2$ (C) 0 (D) $\log 2$

   ›Reveal solutionSolution

   Substituting $t=2^x$ turns the argument into $\frac{2t}{1+t^2}=\sin(2\tan^{-1}t)$, so $f(x)=2\tan^{-1}(2^x)$ and $f'(0)=\log 2$.

   Rewrite the argument with $t=2^x$:

   $\frac{2^{x+1}}{1+4^x}=\frac{2\cdot 2^x}{1+(2^x)^2}=\frac{2t}{1+t^2}.$

   Since $\frac{2t}{1+t^2}=\sin\big(2\tan^{-1}t\big)$, we get

   $f(x)=\sin^{-1}\!\Big(\frac{2t}{1+t^2}\Big)=2\tan^{-1}(2^x).$

   Differentiate: …

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9. COMEDK 2021Set 2021-B1 markMCQ
   Q.If f(1)=3 and f'(1) = 4, then the value of the derivative of $\tan^{-1}[f(x)]$ at x = 1 is (A) 3/17 (B) 4/7 (C) 2/5 (D) 1/2

   ›Reveal solutionSolution

   The derivative at $x=1$ is $\tfrac25$.

   By the chain rule, $\dfrac{d}{dx}\tan^{-1}[f(x)]=\dfrac{f'(x)}{1+[f(x)]^2}$. Substituting $f(1)=3$, $f'(1)=4$: …

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10. COMEDK 2025Set 2025-A1 markMCQ
    Q.Differentiate $\log _a x$ with respect to $a^x$ (A) $\frac{1}{x a^x}$ (B) $\frac{1}{x a^x(\log a)^2}$ (C) $\frac{a^x}{x(\log a)^2}$ (D) $\frac{a^x}{x}$

    ›Reveal solutionSolution

    We want the derivative of $\log_a x$ with respect to $a^x$. Using the chain rule in reverse (differentiating one function of a variable with respect to another function of the same variable), we get $\frac{1}{x a^x (\log a)^2}$, which corresponds to option (B).

    ---

    Concept & Intuition

    The phrase “differentiate $f$ with respect to $g$” means: treat $g$ as the independent variable and find $\frac{df}{dg}$. If both $f$ and $g$ are functions of a common variable (here $x$), we use the chain rule:

    $\frac{df}{dg} = \frac{df/dx}{dg/dx}.$

    So we compute the ordinary derivatives of $\log_a x$ and $a^x$ with respect to $x$, then take their ratio.

    ---

    Step-by-step solution

    1. Rewrite $\log_a x$ in terms of natural logs

    $\log_a x = \frac{\ln x}{\ln a}.$

    This is the standard change-of-base formula. Here $\ln a$ is a constant.

    2. Differentiate $\log_a x$ with respect to $x$

    $\frac{d}{dx} \left( \frac{\ln x}{\ln a} \right) = \frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}.$

    3. Differentiate $a^x$ with respect to $x$

       Recall that $\frac{d}{dx} a^x = a^x \ln a$. (This comes from writing $a^x = e^{x \ln a}$ and using the chain rule.)

    4. Apply the “derivative with respect to” formula

       $$$$

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11. COMEDK 2025Set 2025-M1 markMCQ
    Q.If $f(x)=\left(\frac{3+x}{1+x}\right)^{2+3 x}$, then $f^{\prime}(0)=$ (A) $12+\log 3$ (B) $-12+3 \log 3$ (C) $-\frac{4}{3}+3 \log 3$ (D) $-12+27 \log 3$

    ›Reveal solutionSolution

    To differentiate a function of the form $h(x)^{g(x)}$, we use logarithmic differentiation: take $\log$, differentiate implicitly, then evaluate at $x=0$. The result is $f'(0) = -12 + 27 \log 3$, which corresponds to option (D).

    We have $f(x) = \left( \frac{3+x}{1+x} \right)^{2+3x}$. This is a variable base raised to a variable exponent — a classic case for logarithmic differentiation. The reason: neither the power rule nor the exponential rule alone applies, but taking logs converts the exponent into a factor, letting us use the product rule.

    ---

    1. Take the natural logarithm of both sides Let $y = f(x)$. Then

    $\log y = (2+3x) \log\left( \frac{3+x}{1+x} \right).$

    2. Differentiate implicitly with respect to $x$ On the left: $\frac{d}{dx} \log y = \frac{y'}{y}$. On the right: use the product rule. Let $u = 2+3x$ and $v = \log\left( \frac{3+x}{1+x} \right)$. Then

    $\frac{y'}{y} = u' v + u v'.$

    Here $u' = 3$.

    For $v$, note $\log\left( \frac{3+x}{1+x} \right) = \log(3+x) - \log(1+x)$, so

    $v' = \frac{1}{3+x} - \frac{1}{1+x}.$

    3. Write the derivative expression

    $\frac{y'}{y} = 3 \log\left( \frac{3+x}{1+x} \right) + (2+3x)\left( \frac{1}{3+x} - \frac{1}{1+x} \right).$

    Hence

    $y' = y \left[ 3 \log\left( \frac{3+x}{1+x} \right) + (2+3x)\left( \frac{1}{3+x} - \frac{1}{1+x} \right) \right].$

    4. Evaluate at $x=0$ First, $y(0) = \left( \frac{3}{1} \right)^{2} = 9$. Next, $\log\left( \frac{3+0}{1+0} \right) = \log 3$. …

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12. COMEDK 2026Set 2026-A1 markMCQ
    Q.

    [!FORMULA] $\text { If } y=\tan ^{-1}\left(\frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}-\sqrt{1-x^3}}\right) \text { then } \frac{\boldsymbol{d} \boldsymbol{y}}{\boldsymbol{d x}}=$

    (A) $-\frac{3 x^2}{2 \sqrt{1-x^6}}$ (B) $-\frac{6 x^2}{\sqrt{1-x^6}}$ (C) $\frac{6 x^2}{\sqrt{1-x^6}}$ (D) $\frac{3 x^2}{\sqrt{1-x^6}}$

    ›Reveal solutionSolution

    The key is to simplify the argument of the inverse tangent using the identity $\tan^{-1}\left(\frac{a+b}{a-b}\right) = \frac{\pi}{4} + \tan^{-1}\left(\frac{b}{a}\right)$, then differentiate. The derivative simplifies to $-\frac{3x^2}{2\sqrt{1-x^6}}$, so the correct option is (A).

    We start with

    $y = \tan^{-1}\left(\frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}-\sqrt{1-x^3}}\right).$

    The expression inside looks messy, but there’s a classic trick: when you see a fraction of the form $\frac{A+B}{A-B}$, it often simplifies via the identity

    $\tan^{-1}\left(\frac{A+B}{A-B}\right) = \frac{\pi}{4} + \tan^{-1}\left(\frac{B}{A}\right),$

    provided $A > B > 0$ (which holds here for small $x$). This works because $\tan(\frac{\pi}{4}+\theta) = \frac{1+\tan\theta}{1-\tan\theta}$, and setting $\tan\theta = B/A$ gives exactly our fraction. This reduces the problem to differentiating a much simpler expression.

    1. Apply the identity Let $A = \sqrt{1+x^3}$ and $B = \sqrt{1-x^3}$. Then

    $y = \tan^{-1}\left(\frac{A+B}{A-B}\right) = \frac{\pi}{4} + \tan^{-1}\left(\frac{B}{A}\right).$

    Since $\frac{\pi}{4}$ is constant,

    $\frac{dy}{dx} = \frac{d}{dx} \tan^{-1}\left(\frac{B}{A}\right).$

    2. Simplify the ratio

    $\frac{B}{A} = \frac{\sqrt{1-x^3}}{\sqrt{1+x^3}} = \sqrt{\frac{1-x^3}{1+x^3}}.$

    So

    $y = \frac{\pi}{4} + \tan^{-1}\left( \sqrt{\frac{1-x^3}{1+x^3}} \right).$

    3. Differentiate using the chain rule Let $u = \sqrt{\frac{1-x^3}{1+x^3}}$. Then

    $\frac{dy}{dx} = \frac{1}{1+u^2} \cdot \frac{du}{dx}.$

    First compute $1+u^2$:

    $u^2 = \frac{1-x^3}{1+x^3}, \quad \text{so} \quad 1+u^2 = 1 + \frac{1-x^3}{1+x^3} = \frac{(1+x^3)+(1-x^3)}{1+x^3} = \frac{2}{1+x^3}.$

    Hence

    $\frac{1}{1+u^2} = \frac{1+x^3}{2}.$

    4. Find $\frac{du}{dx}$ Write $u = \left( \frac{1-x^3}{1+x^3} \right)^{1/2}$. Differentiate using the chain rule and quotient rule:

    $\frac{du}{dx} = \frac{1}{2} \left( \frac{1-x^3}{1+x^3} \right)^{-1/2} \cdot \frac{d}{dx}\left( \frac{1-x^3}{1+x^3} \right).$

    The derivative of the quotient:

    $\frac{d}{dx}\left( \frac{1-x^3}{1+x^3} \right) = \frac{(-3x^2)(1+x^3) - (1-x^3)(3x^2)}{(1+x^3)^2} = \frac{-3x^2(1+x^3 + 1 - x^3)}{(1+x^3)^2} = \frac{-6x^2}{(1+x^3)^2}.$

    Also note that

    $$$$

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