Q.Differentiate w.r.t. $x$: $(x + 1)^2 (x + 2)^3 (x + 3)^4$.

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.$\dfrac{d}{dx}\left[\dfrac{(x+1)^2\sqrt{x-1}}{(x+4)^3 e^x}\right] = f(x)\left[\dfrac{2}{x+1} + \dfrac{1}{2(x-1)} - \dfrac{3}{x+4} - 1\right]$, then $f(5) =$ (A) $\dfrac{72}{81}e^5$ (B) $\dfrac{7}{81e^5}$ (C) $\dfrac{8}{81e^5}$ (D) $e^5$

   ›Reveal solutionSolution

   This is logarithmic differentiation in disguise: $f(x)$ is just the original function $y$ itself, so $f(5)$ is obtained by plugging $x=5$ directly into $y$, giving $\dfrac{8}{81e^5}$.

   Concept and Intuition

   Logarithmic differentiation says: if $y$ is a product/quotient of several factors raised to powers, then $\dfrac{1}{y}\dfrac{dy}{dx}$ equals the sum of (power)$\times$(derivative of $\log$ of each factor). The given identity is precisely $\dfrac{dy}{dx}=y\times[\text{sum of log-derivative terms}]$, so the "$f(x)$" multiplying the bracket must be $y$ itself — no extra work is needed beyond recognising the pattern.

   Step-by-Step Solution

   1. Let $y=\dfrac{(x+1)^2\sqrt{x-1}}{(x+4)^3e^x}$.
   2. $\log y=2\log(x+1)+\dfrac12\log(x-1)-3\log(x+4)-x$.
   3. Differentiate both sides w.r.t. $x$: $\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{2}{x+1}+\dfrac{1}{2(x-1)}-\dfrac{3}{x+4}-1$.
   4. So $\dfrac{dy}{dx}=y\left[\dfrac{2}{x+1}+\dfrac{1}{2(x-1)}-\dfrac{3}{x+4}-1\right]$, matching the given form with $f(x)=y(x)$.
   5. Evaluate $y$ at $x=5$: $(x+1)^2=36$, $\sqrt{x-1}=\sqrt4=2$, $(x+4)^3=9^3=729$, $e^x=e^5$. …

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2. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If $y=(x-1)(x+2)(x^2+5)(x^4+8)$, then $\lim_{x\to-1}\left(\dfrac{dy}{dx}\right)=$ (A) $-30$ (B) $30$ (C) $52$ (D) $-52$

   ›Reveal solutionSolution

   Since $y$ is a polynomial, its derivative is continuous, so the limit is just the derivative evaluated at $x=-1$; the product rule for four factors gives $30$.

   Concept and Intuition

   For a product of several differentiable functions, $\left(f_1f_2f_3f_4\right)' = f_1'f_2f_3f_4+f_1f_2'f_3f_4+f_1f_2f_3'f_4+f_1f_2f_3f_4'$ — differentiate one factor at a time, keeping the rest unchanged. Since a polynomial's derivative is itself continuous, $\lim_{x\to-1}y' = y'(-1)$.

   Step-by-Step Solution

   1. Let $f_1=x-1,\ f_2=x+2,\ f_3=x^2+5,\ f_4=x^4+8$, so $f_1'=1,\ f_2'=1,\ f_3'=2x,\ f_4'=4x^3$.
   2. Evaluate all at $x=-1$: $f_1=-2,\ f_2=1,\ f_3=1+5=6,\ f_4=1+8=9$, and $f_3'=-2,\ f_4'=-4$.
   3. Apply the product rule:
      • $f_1'f_2f_3f_4 = (1)(1)(6)(9)=54$
      • $f_1f_2'f_3f_4 = (-2)(1)(6)(9)=-108$ …

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3. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.At $x = \dfrac{\pi^2}{4}$, $\dfrac{d}{dx}\left(\mathrm{Tan}^{-1}(\cos\sqrt{x}) + \mathrm{Sec}^{-1}(e^x)\right) =$ (A) $\dfrac{1}{\sqrt{e^{\pi^2/2} - 1}} - \dfrac{1}{\pi}$ (B) $\dfrac{\pi}{4} + \dfrac{1}{\sqrt{e^{\pi^2} + e^{\pi^2/2}}}$ (C) $\dfrac{1}{\sqrt{e^{\pi^2} + e^{\pi^2/2}}} + \dfrac{2}{\pi}\cot\left(\dfrac{\sqrt{\pi}}{2}\right)$ (D) $\dfrac{1}{\sqrt{e^{\pi}}} + \dfrac{1}{\pi}$

   ›Reveal solutionSolution

   Differentiate each inverse-trig composite separately and evaluate at $x=\pi^2/4$.

   Concept and Intuition

   Use $\frac{d}{dx}\mathrm{Tan}^{-1}(u)=\frac{u'}{1+u^2}$ and $\frac{d}{dx}\mathrm{Sec}^{-1}(u)=\frac{u'}{|u|\sqrt{u^2-1}}$, then substitute the given value of $x$.

   Step-by-Step Solution

   1. Let $u=\cos\sqrt x$. $u'=-\sin(\sqrt x)\cdot\frac{1}{2\sqrt x}$.
   2. At $x=\pi^2/4$: $\sqrt x=\pi/2$, so $\cos(\pi/2)=0\Rightarrow u=0$, and $\sin(\pi/2)=1$, so $u'=-1\cdot\frac{1}{2(\pi/2)}=-\frac1\pi$.
   3. $\frac{d}{dx}\mathrm{Tan}^{-1}(u)=\frac{u'}{1+u^2}=\frac{-1/\pi}{1+0}=-\frac1\pi$.
   4. Now let $v=e^x$. $\frac{d}{dx}\mathrm{Sec}^{-1}(v)=\frac{v'}{|v|\sqrt{v^2-1}}=\frac{e^x}{e^x\sqrt{e^{2x}-1}}=\frac{1}{\sqrt{e^{2x}-1}}$ (using $e^x>0$).
   5. At $x=\pi^2/4$: this is $\frac{1}{\sqrt{e^{\pi^2/2}-1}}$. …

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4. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $y = \log\left(\dfrac{1+x}{1-x}\right)^{1/4} - \dfrac{1}{2}\tan^{-1}(x)$, then $\dfrac{dy}{dx}$ at $x = \dfrac{1}{\sqrt{2}}$ equals ______ (A) $\dfrac{-4}{3}$ (B) $\dfrac{4}{3}$ (C) $\dfrac{-2}{3}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   Tests differentiating a log-of-a-power expression combined with an arctan term, then evaluating at a specific point.

   Concept and Intuition

   Splitting $\log\left[\dfrac{1+x}{1-x}\right]^{1/4}$ using log rules turns it into $\dfrac14[\log(1+x)-\log(1-x)]$, which differentiates term-by-term far more easily than trying to apply the chain rule to the whole power-of-a-quotient directly.

   Step-by-Step Solution

   1. Rewrite $y=\dfrac14\log\left(\dfrac{1+x}{1-x}\right)-\dfrac12\tan^{-1}x = \dfrac14[\log(1+x)-\log(1-x)]-\dfrac12\tan^{-1}x$.
   2. Differentiate: $\dfrac{dy}{dx}=\dfrac14\left[\dfrac{1}{1+x}+\dfrac{1}{1-x}\right]-\dfrac12\cdot\dfrac{1}{1+x^2}$.
   3. Combine the bracket: $\dfrac{1}{1+x}+\dfrac{1}{1-x}=\dfrac{(1-x)+(1+x)}{1-x^2}=\dfrac{2}{1-x^2}$.
   4. So $\dfrac{dy}{dx}=\dfrac14\cdot\dfrac{2}{1-x^2}-\dfrac{1}{2(1+x^2)}=\dfrac{1}{2(1-x^2)}-\dfrac{1}{2(1+x^2)}$.
   5. Combine over a common denominator: $=\dfrac12\cdot\dfrac{(1+x^2)-(1-x^2)}{(1-x^2)(1+x^2)}=\dfrac12\cdot\dfrac{2x^2}{1-x^4}=\dfrac{x^2}{1-x^4}$. …

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5. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Let $f(x) = \log\left(\sqrt[4]{x^5}\right) + \dfrac{e^{-\frac{1}{x}}}{x^3}$. If $f'(x) = \dfrac{e^{-\frac{1}{x}}}{x^3} G(x) + \dfrac{k}{x}$, then the roots of $G(x) + k = 0$ is (A) $\dfrac{3}{4}, 1$ (B) $1, -3$ (C) $\dfrac{2}{5}, 2$ (D) $2, \dfrac{3}{4}$

   ›Reveal solutionSolution

   Differentiate $f(x)$ term by term, match the given form to identify $G(x)$ and $k$, then solve the resulting quadratic $G(x)+k=0$. Roots: $\dfrac25$ and $2$.

   Concept and Intuition

   The function is a sum of a log term and a product term. Differentiating the product term using the product rule and factoring out $\dfrac{e^{-1/x}}{x^3}$ (matching the form given in the problem) isolates $G(x)$ directly by comparison; the log term's derivative isolates $k$.

   Step-by-Step Solution

   1. Rewrite $f(x) = \log\left(x^{5/4}\right) + \dfrac{e^{-1/x}}{x^3} = \dfrac54\log x + e^{-1/x}\,x^{-3}$.
   2. Differentiate the log term: $\dfrac{d}{dx}\left(\dfrac54\log x\right) = \dfrac{5}{4x}$.
   3. Differentiate the product term using the product rule. First, $\dfrac{d}{dx}e^{-1/x} = e^{-1/x}\cdot\dfrac{d}{dx}\left(-\dfrac1x\right) = e^{-1/x}\cdot\dfrac{1}{x^2}$. $\dfrac{d}{dx}\left(e^{-1/x}x^{-3}\right) = \dfrac{e^{-1/x}}{x^2}\cdot x^{-3} + e^{-1/x}\cdot(-3x^{-4}) = e^{-1/x}\left(x^{-5}-3x^{-4}\right)$.
   4. Factor out $\dfrac{e^{-1/x}}{x^3}$: $e^{-1/x}(x^{-5}-3x^{-4}) = \dfrac{e^{-1/x}}{x^3}\left(x^{-2}-3x^{-1}\right) = \dfrac{e^{-1/x}}{x^3}\left(\dfrac{1}{x^2}-\dfrac{3}{x}\right)$. …

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6. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If $\vec{a} = t^2\hat{i} + e^t\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + t^2\hat{j} + \log t\ \hat{k}$ and $f(t) = \vec{a}.\vec{b}$, then $f'(1) =$ (A) $3 + 5e$ (B) $5 + 3e$ (C) $3 + 3e$ (D) $4 + 3e$

   ›Reveal solutionSolution

   Compute the dot product as a scalar function of $t$, differentiate using the product rule, then evaluate at $t=1$ to get $5+3e$.

   Concept and Intuition

   The dot product of two vector functions of $t$ is just a scalar function; differentiate it componentwise like any product of functions, term by term.

   Step-by-Step Solution

   1. $\vec a\cdot\vec b = (t^2)(2) + (e^t)(t^2) + (1)(\log t) = 2t^2 + t^2 e^t + \log t$.
   2. Differentiate: $\frac{d}{dt}(2t^2)=4t$; $\frac{d}{dt}(t^2e^t) = 2te^t+t^2e^t$ (product rule); $\frac{d}{dt}(\log t)=\frac1t$. …

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7. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.If $y = \left(1 + \dfrac{1}{x}\right)\left(1 + \dfrac{2}{x}\right)\left(1 + \dfrac{3}{x}\right)\cdots\cdots\left(1 + \dfrac{n}{x}\right)$ and $x \ne 0$. When $x = -1$, $\dfrac{dy}{dx} =$ _______ (A) $n!$ (B) $(n-1)!$ (C) $(-1)^n (n-1)!$ (D) $(-1)^n n!$

   ›Reveal solutionSolution

   This tests differentiating a product where one factor vanishes exactly at the evaluation point, so only one product-rule term survives. Answer: $(-1)^n(n-1)!$.

   Concept and Intuition

   When $y = f_1(x)g(x)$ and $f_1(x_0)=0$, then $y'(x_0) = f_1'(x_0)g(x_0) + f_1(x_0)g'(x_0) = f_1'(x_0)g(x_0)$ — the second term drops out automatically. Here $f_1(x)=1+\tfrac1x$ vanishes precisely at $x=-1$, which simplifies the whole problem.

   Step-by-Step Solution

   1. Write $y = \left(1+\tfrac1x\right)g(x)$ with $g(x) = \prod_{k=2}^{n}\left(1+\tfrac kx\right)$.
   2. $f_1(x) = 1+\tfrac1x$, so $f_1(-1) = 1-1 = 0$ — confirming the split is valid and useful.
   3. $f_1'(x) = -\tfrac{1}{x^2}$, so $f_1'(-1) = -1$.
   4. $g(-1) = \prod_{k=2}^{n}\left(1-k\right) = (-1)(-2)\cdots(-(n-1)) = (-1)^{n-1}(n-1)!$. …

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8. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If $f(x)=5\cos^3x-3\sin^2x$ and $g(x)=4\sin^3x+\cos^2x$, then the derivative of $f(x)$ with respect to $g(x)$ is (A) $\dfrac{5\cos x+2}{6\cos x-1}$ (B) $-\left(\dfrac{5\cos x+2}{6\cos x-1}\right)$ (C) $\dfrac{15\cos x-6}{12\sin x+2}$ (D) $-\left(\dfrac{15\cos x+6}{12\sin x-2}\right)$

   ›Reveal solutionSolution

   The derivative of one function with respect to another is the ratio of their derivatives w.r.t. $x$; simplifying gives $-\dfrac{15\cos x+6}{12\sin x-2}$.

   Concept and Intuition

   When asked for $\dfrac{df}{dg}$ (derivative of $f$ with respect to $g$, not $x$), the chain rule gives $\dfrac{df}{dg}=\dfrac{df/dx}{dg/dx}$, valid wherever $g'(x)\ne0$. So computing both ordinary derivatives w.r.t. $x$ and dividing solves it directly.

   Step-by-Step Solution

   1. $f(x)=5\cos^3x-3\sin^2x$. $f'(x)=5\cdot3\cos^2x(-\sin x)-3\cdot2\sin x\cos x=-15\cos^2x\sin x-6\sin x\cos x=-3\sin x\cos x(5\cos x+2)$.
   2. $g(x)=4\sin^3x+\cos^2x$. $g'(x)=4\cdot3\sin^2x\cos x+2\cos x(-\sin x)=12\sin^2x\cos x-2\sin x\cos x=2\sin x\cos x(6\sin x-1)$. …

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9. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If $y = \sqrt{\dfrac{x^4\sqrt{3x-5}}{(x^2-3)(2x-3)}}$, then $\left(\dfrac{dy}{dx}\right)_{x=2} =$ (A) $5$ (B) $0$ (C) $1$ (D) $-5$

   ›Reveal solutionSolution

   Logarithmic differentiation converts the messy root/product/quotient into a sum of logs; evaluating at $x=2$ gives $y'=-5$.

   Concept and Intuition

   Whenever a function is a product, quotient, or power of several simpler factors (especially under a square root), logarithmic differentiation is the cleanest route: take $\log$ of both sides to convert products to sums and powers to multiples, differentiate termwise, then multiply back by $y$.

   Step-by-Step Solution

   1. Write $y=\left[\dfrac{x^4\sqrt{3x-5}}{(x^2-3)(2x-3)}\right]^{1/2}$. Taking $\log$:

   $\log y = \frac12\Big[4\log x + \frac12\log(3x-5) - \log(x^2-3) - \log(2x-3)\Big].$

   2. Differentiate both sides with respect to $x$:

   $\frac{y'}{y}=\frac12\left[\frac{4}{x}+\frac12\cdot\frac{3}{3x-5}-\frac{2x}{x^2-3}-\frac{2}{2x-3}\right].$

   3. Evaluate the original $y$ at $x=2$: $x^4=16$, $3x-5=1\Rightarrow\sqrt{1}=1$, $x^2-3=1$, $2x-3=1$. So $y=\sqrt{16\cdot1/(1\cdot1)}=\sqrt{16}=4$.
   4. Evaluate each bracket term at $x=2$:
      • $\dfrac{4}{x}=\dfrac{4}{2}=2$
      • $\dfrac12\cdot\dfrac{3}{3x-5}=\dfrac12\cdot\dfrac{3}{1}=1.5$ …

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10. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.If $f(x) = 2x^2 + 3x - 5$, then the value of $f'(0) + 3f'(-1)$ is equal to _______ (A) 1 (B) 0 (C) 3 (D) 2

    ›Reveal solutionSolution

    This tests basic polynomial differentiation and evaluation at points. Answer: $0$.

    Concept and Intuition

    Differentiate the polynomial term by term using the power rule, then substitute the given values directly.

    Step-by-Step Solution

    1. $f(x)=2x^2+3x-5 \Rightarrow f'(x) = 4x+3$.
    2. $f'(0) = 4(0)+3 = 3$.
    3. $f'(-1) = 4(-1)+3 = -4+3=-1$. …

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11. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.If $y=\mathrm{Tan}^{-1}\left(\dfrac{1}{1+x+x^2}\right)+\mathrm{Tan}^{-1}\left(\dfrac{1}{x^2+3x+3}\right)+\mathrm{Tan}^{-1}\left(\dfrac{1}{x^2+5x+7}\right)$, then $y'(0)=$ (A) $-\dfrac{3}{10}$ (B) $-\dfrac{1}{2}$ (C) $-\dfrac{7}{10}$ (D) $-\dfrac{9}{10}$

    ›Reveal solutionSolution

    Each $\arctan$ term telescopes using $\arctan A - \arctan B=\arctan\frac{A-B}{1+AB}$, collapsing the whole sum to $\arctan(x+3)-\arctan x$; differentiating and plugging $x=0$ gives $-9/10$.

    Concept and Intuition

    The identity $\arctan A-\arctan B=\arctan\dfrac{A-B}{1+AB}$ (when $AB>-1$) is the key: if we can write each denominator $1+x+x^2$ etc. as $1+AB$ for consecutive integers-shifted $A,B$ with $A-B=1$, the arctan of the reciprocal collapses to a difference of two arctans. Stacking three such differences telescopes almost everything away, leaving only the first and last terms.

    Step-by-Step Solution

    1. First term: want $A-B=1,\ AB=x+x^2=x(x+1)$. Take $A=x+1,B=x$: $A-B=1$ ✓, $AB=x(x+1)=x^2+x$ ✓. So $\arctan\dfrac{1}{1+x+x^2}=\arctan(x+1)-\arctan x$.
    2. Second term: want $AB=x^2+3x+2=(x+1)(x+2)$. Take $A=x+2,B=x+1$: $AB=(x+1)(x+2)$ ✓. So $\arctan\dfrac{1}{x^2+3x+3}=\arctan(x+2)-\arctan(x+1)$.
    3. Third term: want $AB=x^2+5x+6=(x+2)(x+3)$. Take $A=x+3,B=x+2$. So $\arctan\dfrac{1}{x^2+5x+7}=\arctan(x+3)-\arctan(x+2)$.
    4. Sum: $y=[\arctan(x+1)-\arctan x]+[\arctan(x+2)-\arctan(x+1)]+[\arctan(x+3)-\arctan(x+2)]$. …

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12. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.$f(x) = \sqrt{x^2+1}$; $g(x) = \dfrac{x+1}{x^2+1}$; $h(x) = 2x - 3$. Then the value of $f'[h'(g'(x))] =$ (A) $\sqrt{5}$ (B) $\dfrac{2}{\sqrt{5}}$ (C) $\dfrac{\sqrt{5}}{2}$ (D) $\dfrac{1}{\sqrt{5}}$

    ›Reveal solutionSolution

    Because $h(x)=2x-3$ is linear, $h'$ is the constant function $2$ — so the whole nested expression collapses to $f'(2)=2/\sqrt5$, and $g$'s formula is a red herring.

    Concept and Intuition

    The expression $f'[h'(g'(x))]$ looks like it needs $g'(x)$ computed explicitly and plugged through two more functions. But $h(x)=2x-3$ is linear, so its derivative $h'(x)=2$ for every input — $h'$ doesn't actually depend on its argument at all. That means whatever $g'(x)$ evaluates to, feeding it into $h'$ always returns $2$. The problem is testing whether you notice this shortcut rather than grinding through $g'(x)$ unnecessarily.

    Step-by-Step Solution

    1. $h(x)=2x-3 \Rightarrow h'(x)=2$ for all $x$ — a constant function.
    2. Therefore $h'(g'(x))=2$, whatever $g'(x)$ actually is (no need to compute it).
    3. So $f'[h'(g'(x))]=f'(2)$. …

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