Q.If 𝑓(𝑥) = 𝑥 tan−1 𝑥 , then 𝑓′(1)is equal to
(A) 𝜋 4 − 1 2
(B) 𝜋 4 + 1 2
(C) − 𝜋 4 − 1 2
(D) − 𝜋 4 + 1 2

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $f(x) = \mathrm{Tan}^{-1}\left(\frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right)$, then $f'\left(-\frac{1}{2}\right) =$ (A) $-\frac{1}{\sqrt{2}}$ (B) $\frac{1}{\sqrt{2}}$ (C) $-\frac{2}{\sqrt{15}}$ (D) $\frac{2}{\sqrt{15}}$

   ›Reveal solutionSolution

   This tests simplifying an inverse-trig expression via a trigonometric substitution before differentiating. Once simplified, $f(x)=\frac\pi4+\frac12\cos^{-1}(x^2)$, giving $f'\left(-\frac12\right)=\frac2{\sqrt{15}}$.

   Concept and Intuition

   Expressions with $\sqrt{1+x^2}$ and $\sqrt{1-x^2}$ together strongly suggest substituting $x^2=\cos\varphi$, turning both square roots into half-angle sine/cosine forms via $1\pm\cos\varphi=2\cos^2\frac\varphi2$ or $2\sin^2\frac\varphi2$. This collapses the arctangent of a ratio into a simple tangent addition, making the function (and its derivative) far easier to handle than direct differentiation of the original expression.

   Step-by-Step Solution

   1. Let $x^2=\cos\varphi$. Then $1+x^2=1+\cos\varphi=2\cos^2\frac\varphi2$ and $1-x^2=1-\cos\varphi=2\sin^2\frac\varphi2$.
   2. So $\sqrt{1+x^2}=\sqrt2\cos\frac\varphi2$ and $\sqrt{1-x^2}=\sqrt2\sin\frac\varphi2$ (taking the principal positive roots).
   3. The ratio inside $f$: $\frac{\cos\frac\varphi2+\sin\frac\varphi2}{\cos\frac\varphi2-\sin\frac\varphi2}=\frac{1+\tan\frac\varphi2}{1-\tan\frac\varphi2}=\tan\left(\frac\pi4+\frac\varphi2\right)$.
   4. So $f(x)=\tan^{-1}\left[\tan\left(\frac\pi4+\frac\varphi2\right)\right]=\frac\pi4+\frac\varphi2=\frac\pi4+\frac12\cos^{-1}(x^2)$. …

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2. 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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3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $f(x) = \sqrt{-(1 + x)}\ \mathrm{Sec}^{-1}x$ is a real valued function, then $f'(x) =$ (A) $-\dfrac{\mathrm{Sec}^{-1}x}{2\sqrt{-(1+x)}} + \dfrac{1}{x\sqrt{x-1}}$ (B) $-\dfrac{\mathrm{Sec}^{-1}x}{2\sqrt{-(1+x)}} - \dfrac{1}{x\sqrt{1-x}}$ (C) $-\dfrac{\mathrm{Sec}^{-1}x}{2\sqrt{-(1+x)}} - \dfrac{1}{x\sqrt{x-1}}$ (D) $-\dfrac{\mathrm{Sec}^{-1}x}{2\sqrt{-(1+x)}} + \dfrac{1}{x\sqrt{1-x}}$

   ›Reveal solutionSolution

   This tests differentiating a product involving $\mathrm{Sec}^{-1}x$ on its $x\le-1$ branch, being careful with the sign of $|x|$ inside the arcsec derivative; the answer is option (B).

   Concept and Intuition

   $f$ is only real for $x\le-1$ (so that $-(1+x)\ge0$ and $|x|\ge1$). On this branch, $\sqrt{x^2-1}$ can be split as $\sqrt{1-x}\cdot\sqrt{-1-x}$ — both factors positive when $x\le-1$ — and $\sqrt{-1-x}$ is exactly the $u$ already in the problem, which lets the answer be written in the given form.

   Step-by-Step Solution

   1. Let $u=\sqrt{-(1+x)}$, $v=\mathrm{Sec}^{-1}x$, so $f=uv$ and $f'=u'v+uv'$.
   2. $u' = \dfrac{1}{2\sqrt{-(1+x)}}\cdot(-1) = -\dfrac{1}{2u}$.
   3. Standard result: $\dfrac{d}{dx}\mathrm{Sec}^{-1}x = \dfrac{1}{|x|\sqrt{x^2-1}}$. For $x\le-1$, $|x|=-x$, so $v' = -\dfrac{1}{x\sqrt{x^2-1}}$.
   4. For $x\le-1$: $\sqrt{x^2-1}=\sqrt{(1-x)(-1-x)}=\sqrt{1-x}\cdot\sqrt{-1-x}=\sqrt{1-x}\cdot u$.
   5. So $v' = -\dfrac{1}{x\sqrt{1-x}\,u}$. …

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4. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.The derivative of $\text{Sec}^{-1}\left(\dfrac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\dfrac{1}{2}$ equals ________ (A) 2 (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{4}$ (D) 4

   ›Reveal solutionSolution

   Substituting $x=\cos\theta$ reduces both functions to simple multiples of $\theta$, giving the derivative of one with respect to the other as $2/x$, which equals $4$ at $x=1/2$.

   Concept and Intuition

   When asked for the derivative of one function of $x$ with respect to another function of $x$ (parametric-style differentiation), the trick is $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}$. Here, substituting $x=\cos\theta$ turns the ugly $\operatorname{Sec}^{-1}\!\left(\dfrac{1}{2x^2-1}\right)$ into the clean double-angle expression $2\theta$.

   Step-by-Step Solution

   1. Let $x=\cos\theta$, $\theta\in[0,\pi]$, so $\sin\theta=\sqrt{1-x^2}\ge0$.
   2. Then $2x^2-1 = 2\cos^2\theta-1=\cos2\theta$, so $\dfrac{1}{2x^2-1}=\sec2\theta$.
   3. So $u=\operatorname{Sec}^{-1}(\sec2\theta)=2\theta = 2\operatorname{Cos}^{-1}x$ (valid since at $x=1/2$, $2\theta=2\pi/3\in[0,\pi]\setminus\{\pi/2\}$, the correct principal range for $\operatorname{Sec}^{-1}$).
   4. $\dfrac{du}{dx} = 2\cdot\left(\dfrac{-1}{\sqrt{1-x^2}}\right) = \dfrac{-2}{\sqrt{1-x^2}}$.
   5. $v=\sqrt{1-x^2}$, so $\dfrac{dv}{dx} = \dfrac{-x}{\sqrt{1-x^2}}$. …

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5. 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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6. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.$\dfrac{d}{dx}\left\{\sin^2\left(\mathrm{Cot}^{-1}\sqrt{\dfrac{1+x}{1-x}}\right)\right\} =$ (A) 0 (B) $\dfrac{1}{2}$ (C) $\dfrac{-1}{2}$ (D) $-1$

   ›Reveal solutionSolution

   This tests simplifying an inverse-trig composite using the identity $\sin^2\theta=\tfrac{1}{1+\cot^2\theta}$ before differentiating, avoiding messy chain-rule work. Answer: $-\tfrac12$.

   Concept and Intuition

   Rather than differentiating $\sin^2(\mathrm{Cot}^{-1}(\cdots))$ directly through the chain rule, it's far simpler to algebraically simplify the whole expression to a function of $x$ first, since $\cot\theta$ is given explicitly.

   Step-by-Step Solution

   1. Let $\theta = \mathrm{Cot}^{-1}\sqrt{\dfrac{1+x}{1-x}}$, so $\cot\theta = \sqrt{\dfrac{1+x}{1-x}}$, hence $\cot^2\theta = \dfrac{1+x}{1-x}$.
   2. Using $\sin^2\theta = \dfrac{1}{1+\cot^2\theta}$: $\sin^2\theta = \dfrac{1}{1+\frac{1+x}{1-x}} = \dfrac{1-x}{(1-x)+(1+x)} = \dfrac{1-x}{2}$. …

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7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The derivative of $\mathrm{Sec}^{-1}\left(\dfrac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\dfrac{1}{2}$ is (A) $-2$ (B) $1$ (C) $2$ (D) $4$

   ›Reveal solutionSolution

   A "derivative with respect to another function" problem — the substitution $x=\cos\theta$ collapses $\mathrm{Sec}^{-1}(2x^2-1)^{-1}$ into $2\theta$, making the ratio of derivatives trivial. Answer: $4$.

   Concept and Intuition

   To find $\dfrac{du}{dv}$ where $u=u(x)$ and $v=v(x)$, use $\dfrac{du}{dv}=\dfrac{du/dx}{dv/dx}$ (or, more cleanly here, a common parameter $\theta$). The key trick recognizing $2x^2-1$ as $\cos2\theta$ when $x=\cos\theta$ turns the inverse secant of a rational expression into a simple linear function of $\theta$.

   Step-by-Step Solution

   1. Let $x=\cos\theta$, $\theta\in[0,\pi]$ (the natural domain for $\cos^{-1}$).
   2. Then $2x^2-1 = 2\cos^2\theta-1=\cos2\theta$, so $\dfrac{1}{2x^2-1}=\dfrac{1}{\cos2\theta}=\sec2\theta$.
   3. Hence $u=\mathrm{Sec}^{-1}(\sec 2\theta)$. Since $x=\tfrac12\Rightarrow\theta=\cos^{-1}\!\left(\tfrac12\right)=\tfrac{\pi}{3}$, we get $2\theta=\tfrac{2\pi}{3}$, which lies in $[0,\pi]$ — the principal range of $\mathrm{Sec}^{-1}$ — so $u=2\theta=2\cos^{-1}x$ validly (no branch correction needed here).
   4. Also $v=\sqrt{1-x^2}=\sqrt{1-\cos^2\theta}=\sin\theta$ (non-negative since $\theta\in[0,\pi]$).
   5. Differentiate w.r.t. $\theta$: $\dfrac{du}{d\theta}=2$, $\dfrac{dv}{d\theta}=\cos\theta=x$.
   6. So $\dfrac{du}{dv}=\dfrac{du/d\theta}{dv/d\theta}=\dfrac{2}{x}$.
   7. At $x=\dfrac12$: $\dfrac{du}{dv}=\dfrac{2}{1/2}=4$. …

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8. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.If $f(t)=\dfrac{t}{2}+\dfrac{1}{4}\log(2t-1)$, then $f'\left(\dfrac{t+1}{2t+1}\right)=$ (A) t (B) 1+t (C) 2t+1 (D) t-1

   ›Reveal solutionSolution

   Differentiate $f(t)$ once, then substitute the given argument; the algebra telescopes neatly because $2s-1$ turns out to be the reciprocal of $2t+1$, giving $f'(s)=t+1$.

   Concept and Intuition

   This is a "plug an expression into a known derivative" problem — compute $f'$ symbolically first, then substitute. The key simplification is noticing that $2\left(\dfrac{t+1}{2t+1}\right)-1$ collapses to $\dfrac{1}{2t+1}$.

   Step-by-Step Solution

   1. $f(t)=\dfrac t2+\dfrac14\log(2t-1)\Rightarrow f'(t)=\dfrac12+\dfrac14\cdot\dfrac{2}{2t-1}=\dfrac12+\dfrac{1}{2(2t-1)}$.
   2. Let $s=\dfrac{t+1}{2t+1}$. Compute $2s-1=\dfrac{2(t+1)-(2t+1)}{2t+1}=\dfrac{1}{2t+1}$.
   3. So $\dfrac{1}{2(2s-1)}=\dfrac{2t+1}{2}$. …

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9. 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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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 2025Set eng-2025-05-24-FN1 markMCQ
    Q.If $y = \mathrm{Tan}^{-1}\left(\dfrac{3x - x^3}{1 - 3x^2}\right) + \mathrm{Tan}^{-1}\left(\dfrac{7x}{1 - 12x^2}\right)$, then at $x = 0$, $\dfrac{dy}{dx} =$ (A) $6$ (B) $7$ (C) $9$ (D) $10$

    ›Reveal solutionSolution

    Differentiate the sum of two arctangent expressions term-by-term and evaluate at $x=0$; the answer is $10$.

    Concept and Intuition

    Rather than trying to recognize the whole expression as some multiple-angle identity, it is safest and fastest to differentiate each $\mathrm{Tan}^{-1}(\cdot)$ term directly using the chain rule $\dfrac{d}{dx}\mathrm{Tan}^{-1}(u)=\dfrac{u'}{1+u^2}$, then substitute $x=0$. Since we only need the derivative at a point, we don't need the general antiderivative simplification (e.g., recognizing $\frac{3x-x^3}{1-3x^2}$ as $\tan(3\theta)$ for $x=\tan\theta$) — direct differentiation is more robust.

    Step-by-Step Solution

    1. Let $y = \mathrm{Tan}^{-1}(u) + \mathrm{Tan}^{-1}(v)$ where $u=\dfrac{3x-x^3}{1-3x^2}$ and $v=\dfrac{7x}{1-12x^2}$.
    2. First term derivative at $x=0$:

    $u'=\frac{(3-3x^2)(1-3x^2)-(3x-x^3)(-6x)}{(1-3x^2)^2}.$

    At $x=0$: numerator $=(3)(1)-0=3$, denominator $=1$, so $u'(0)=3$. Also $u(0)=0$.

    Contribution: $\dfrac{u'(0)}{1+u(0)^2}=\dfrac{3}{1}=3$.

    3. Second term derivative at $x=0$:

    $v'=\frac{7(1-12x^2)-7x(-24x)}{(1-12x^2)^2}.$

    At $x=0$: numerator $=7(1)-0=7$, denominator $=1$, so $v'(0)=7$. Also $v(0)=0$. …

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12. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.Given that $\frac{d}{dx}\int_0^{\phi(x)}f(t)dt=f(\phi(x))\phi'(x)$. For all $x\in\left(0,\frac{\pi}{2}\right)$, if $\int_1^{\cos x}t^2f(t)dt=\cos 2x$, then $f\left(\frac{1}{\sqrt2}\right)=$ (A) $2\sqrt2$ (B) $4\sqrt2$ (C) $\frac{\pi}{4}$ (D) $\frac{-\pi}{4}$

    ›Reveal solutionSolution

    Differentiating the given integral identity using the Leibniz rule for a variable upper limit yields $f(\cos x)=4/\cos x$, so $f(1/\sqrt2)=4\sqrt2$.

    Concept and Intuition

    The stated rule $\frac{d}{dx}\int_0^{\phi(x)}f(t)dt=f(\phi(x))\phi'(x)$ is just the chain rule applied to the Fundamental Theorem of Calculus; it extends immediately to any constant lower limit and any integrand (here $t^2f(t)$ instead of $f(t)$). Differentiating both sides of a functional identity is the standard trick for extracting the value of $f$ at a specific point from an integral equation.

    Step-by-Step Solution

    1. Differentiate $\int_1^{\cos x}t^2f(t)\,dt=\cos2x$ with respect to $x$.
    2. LHS: by the Leibniz rule (with $\phi(x)=\cos x$, $g(t)=t^2f(t)$), $\dfrac{d}{dx}\int_1^{\cos x}t^2f(t)dt=(\cos x)^2f(\cos x)\cdot(-\sin x)$.
    3. RHS: $\dfrac{d}{dx}\cos2x=-2\sin2x=-4\sin x\cos x$.
    4. Equate: $-\sin x\cos^2x\,f(\cos x)=-4\sin x\cos x$. …

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