Q.Find the derivative of $f(x) = 3$ at $x = 0$ and at $x = 3$.

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\displaystyle\lim_{x \to e} \dfrac{(\log x)^{\tan x} - 1}{x - e} =$ (A) $\dfrac{\tan e}{e}$ (B) $\dfrac{1}{\tan e}$ (C) $\dfrac{1}{e}$ (D) $\dfrac{e}{\tan e}$

   ›Reveal solutionSolution

   The limit is the derivative of $f(x)=(\log x)^{\tan x}$ at $x=e$, found by logarithmic differentiation; the answer is $\dfrac{\tan e}{e}$.

   Concept and Intuition

   The expression $\dfrac{f(x)-f(e)}{x-e}$ as $x\to e$ is literally the definition of $f'(e)$, provided $f(e)$ equals the constant subtracted (here $1$). Since $(\log e)^{\tan e}=1^{\tan e}=1$, the numerator $(\log x)^{\tan x}-1$ is exactly $f(x)-f(e)$. So instead of manipulating a $1^\infty$-type limit with tricks, we can simply differentiate $f$ and evaluate at $e$ — logarithmic differentiation is the natural tool because the base and exponent both vary with $x$.

   Step-by-Step Solution

   1. Recognise the limit as $f'(e)$ for $f(x)=(\log x)^{\tan x}$, since $f(e)=1^{\tan e}=1$.
   2. Take logs: $\log f(x) = \tan x\cdot \log(\log x)$.
   3. Differentiate both sides w.r.t. $x$: $\dfrac{f'(x)}{f(x)} = \sec^2 x\cdot \log(\log x) + \tan x\cdot \dfrac{1}{\log x}\cdot\dfrac{1}{x}$.
   4. Evaluate at $x=e$: $\log x = 1 \Rightarrow \log(\log x)=\log 1 = 0$, so the first term vanishes. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $f(x) = \begin{cases} \dfrac{1 - \sin^3 x}{3\cos^2 x}, & x \neq \pi/2 \\ \dfrac{1}{2}, & x = \pi/2 \end{cases}$, then $f'(\pi/2) =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $-1$ (D) $0$

   ›Reveal solutionSolution

   The $0/0$-looking piecewise definition simplifies to a genuinely smooth function near

   $x=\pi/2$; differentiating that smooth form shows every term vanishes at $\pi/2$

   because of a $\cos x$ factor, so $f'(\pi/2)=0$.

   Concept and Intuition

   A piecewise function defined by a "removable" formula plus a patched value is

   differentiable at the patch point exactly when the algebraically simplified (single)

   formula is smooth there and agrees with the patched value — then the derivative of the

   simplified formula IS the derivative of $f$.

   Step-by-Step Solution

   1. Factor: $1-\sin^3x=(1-\sin x)(1+\sin x+\sin^2x)$, and $3\cos^2x=3(1-\sin x)(1+\sin x)$.
   2. For $x\ne\pi/2$: $f(x)=\dfrac{(1-\sin x)(1+\sin x+\sin^2x)}{3(1-\sin x)(1+\sin x)} =\dfrac{1+\sin x+\sin^2x}{3(1+\sin x)}=g(x)$.
   3. Check continuity: $g(\pi/2)=\dfrac{1+1+1}{3(2)}=\dfrac12=f(\pi/2)$ — matches the given patch value exactly, so $f\equiv g$ in a full neighbourhood of $\pi/2$ (not just near it), and $f'(\pi/2)=g'(\pi/2)$.
   4. Differentiate $g(x)=\dfrac{1+\sin x+\sin^2x}{3(1+\sin x)}$ using the quotient rule with $s=\sin x,\ c=\cos x$: numerator of $g'$ works out to $3c\big[(1+2s)(1+s)-(1+s+s^2)\big]=3c(2s+s^2)$, so …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\lim_{x \to e} \dfrac{\log x - 1}{x - e} =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{e}$ (D) Does not exist

   ›Reveal solutionSolution

   This tests L'Hôpital's Rule on a $0/0$ indeterminate form built around $\log e = 1$. The answer is $\dfrac{1}{e}$.

   Concept and Intuition

   As $x \to e$, both $\log x - 1 \to 0$ (because $\log e = 1$) and $x - e \to 0$, so we have a genuine $0/0$ indeterminate form — direct substitution fails and we need L'Hôpital's Rule (differentiate top and bottom separately and re-take the limit), which works because both functions are differentiable near $x=e$ with a nonzero denominator derivative.

   Step-by-Step Solution

   1. Check the form: at $x=e$, numerator $=\log e - 1 = 1-1=0$; denominator $=e-e=0$. So it's $0/0$.
   2. Differentiate numerator: $\dfrac{d}{dx}(\log x - 1) = \dfrac{1}{x}$.
   3. Differentiate denominator: $\dfrac{d}{dx}(x-e) = 1$.
   4. By L'Hôpital's Rule, $\displaystyle\lim_{x\to e}\frac{\log x - 1}{x-e} = \lim_{x\to e}\frac{1/x}{1} = \frac{1}{e}$.

   Common Mistakes …

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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $x \in [-1, 1]$ and $y = (\mathrm{Cot}^{-1}x)^{\mathrm{Cos}^{-1}x}$, then $\left(\dfrac{dy}{dx}\right)_{x=0} =$ (A) $\left(\dfrac{\pi}{2}\right)^{\frac{\pi}{2}}\left(1 + \log\dfrac{\pi}{2}\right)$ (B) $\left(\dfrac{\pi}{2}\right)^{\frac{\pi}{2}}\left(1 - \log\dfrac{\pi}{2}\right)$ (C) $-\left(\dfrac{\pi}{2}\right)^{\frac{\pi}{2}}\left(1 + \log\dfrac{\pi}{2}\right)$ (D) $\left(\dfrac{\pi}{2}\right)^{\frac{\pi}{2}}\left(\log\dfrac{\pi}{2} - 1\right)$

   ›Reveal solutionSolution

   This tests logarithmic differentiation of a function-to-the-function-power form $u(x)^{v(x)}$; evaluated at $x=0$ the derivative is $-\left(\frac\pi2\right)^{\pi/2}\left(1+\log\frac\pi2\right)$.

   Concept and Intuition

   For $y=u^v$ where both $u$ and $v$ are functions of $x$, take logs first ($\log y = v\log u$) and differentiate implicitly — this avoids trying to apply the power rule or exponential rule alone, neither of which applies when both the base and the exponent vary.

   Step-by-Step Solution

   1. Let $u=\mathrm{Cot}^{-1}x$, $v=\mathrm{Cos}^{-1}x$, so $y=u^v$ and $\log y = v\log u$.
   2. Differentiating: $\dfrac{y'}{y} = v'\log u + v\cdot\dfrac{u'}{u}$.
   3. $u' = \dfrac{d}{dx}\mathrm{Cot}^{-1}x = -\dfrac{1}{1+x^2}$; $v' = \dfrac{d}{dx}\mathrm{Cos}^{-1}x = -\dfrac{1}{\sqrt{1-x^2}}$.
   4. At $x=0$: $u(0)=\mathrm{Cot}^{-1}0=\dfrac\pi2$, $v(0)=\mathrm{Cos}^{-1}0=\dfrac\pi2$, $u'(0)=-1$, $v'(0)=-1$.
   5. $\dfrac{y'(0)}{y(0)} = (-1)\log\dfrac\pi2 + \dfrac\pi2\cdot\dfrac{-1}{\pi/2} = -\log\dfrac\pi2 - 1 = -\left(1+\log\dfrac\pi2\right)$. …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $f(x) = \mathrm{Cot}^{-1}\left(\sqrt{\cos 2x}\right)$, then $f'\left(\dfrac{\pi}{6}\right) =$ (A) $\dfrac{1}{\sqrt{3}}$ (B) $\dfrac{2}{\sqrt{3}}$ (C) $\sqrt{\dfrac{2}{3}}$ (D) $\dfrac{-2}{\sqrt{3}}$

   ›Reveal solutionSolution

   Differentiating $\cot^{-1}(\sqrt{\cos2x})$ simplifies neatly to $f'(x)=\tan x/\sqrt{\cos2x}$;

   evaluating at $x=\pi/6$ gives $\sqrt{2/3}$.

   Concept and Intuition

   Composite inverse-trig derivatives often simplify dramatically if you keep the algebra symbolic

   as long as possible and only substitute the specific angle at the very end — and if you notice

   useful trig identities like $1+\cos2x=2\cos^2x$ along the way.

   Step-by-Step Solution

   1. Let $u=\sqrt{\cos2x}$, so $f(x)=\cot^{-1}u$, and $\dfrac{d}{du}\cot^{-1}u=\dfrac{-1}{1+u^2}$.
   2. $u'=\dfrac{1}{2\sqrt{\cos2x}}\cdot(-2\sin2x)=\dfrac{-\sin2x}{\sqrt{\cos2x}}$.
   3. $1+u^2 = 1+\cos2x = 2\cos^2x$.
   4. Chain rule:

   $f'(x) = \frac{-u'}{1+u^2} = \frac{-1}{2\cos^2x}\cdot\left(\frac{-\sin2x}{\sqrt{\cos2x}}\right) = \frac{\sin2x}{2\cos^2x\sqrt{\cos2x}}$

   5. Since $\sin2x=2\sin x\cos x$:

   $\frac{\sin2x}{2\cos^2x} = \frac{2\sin x\cos x}{2\cos^2x}=\tan x$

   so $f'(x)=\dfrac{\tan x}{\sqrt{\cos2x}}$.

   6. At $x=\dfrac{\pi}{6}$: $\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3}$, and

   $\cos\dfrac{\pi}{3}=\dfrac12 \Rightarrow \sqrt{\cos2x}=\dfrac{1}{\sqrt2}$. …

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6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If $y=\log(\sec(\tan^{-1}x))\ (x>0)$, then $\dfrac{dy}{dx}$ at $x=1$ is (A) 1 (B) 3 (C) $\dfrac{1}{2}$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   Simplify $\sec(\tan^{-1}x)=\sqrt{1+x^2}$ first, turning $y$ into an elementary log function; the derivative at $x=1$ is $\dfrac12$.

   Concept and Intuition

   Rather than differentiating $\log(\sec(\tan^{-1}x))$ directly with nested chain rules, it pays to simplify the composition first: if $\theta=\tan^{-1}x$, then in a right triangle with opposite $x$ and adjacent $1$, the hypotenuse is $\sqrt{1+x^2}$, so $\sec\theta=\sqrt{1+x^2}$. This collapses the whole expression to $y=\tfrac12\log(1+x^2)$, an easy function to differentiate.

   Step-by-Step Solution

   1. Let $\theta=\tan^{-1}x$. Then $\tan\theta=x$, and since $\theta\in\left(-\tfrac\pi2,\tfrac\pi2\right)$, $\sec\theta>0$.
   2. $\sec^2\theta=1+\tan^2\theta=1+x^2 \Rightarrow \sec\theta=\sqrt{1+x^2}$.
   3. $y=\log(\sec\theta)=\log\sqrt{1+x^2}=\dfrac12\log(1+x^2)$. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $y = (\log x)^{1/x} + x^{\log x}$, at $x = e$, $\dfrac{dy}{dx} =$ (A) $2 + \dfrac{1}{e}$ (B) $e^2 + \dfrac{1}{2}$ (C) $\dfrac{1}{e^2} + 2$ (D) $e + \dfrac{1}{e}$

   ›Reveal solutionSolution

   Both terms are variable-to-variable-power expressions; use logarithmic

   differentiation on each separately, then plug in $x=e$ (which conveniently

   makes $\log x=1$, killing the messy $\log(\log x)$ term).

   Concept and Intuition

   Expressions of the form (function of $x$)$^{\text{(function of }x)}$ can't be

   differentiated by the power rule or exponential rule alone — logarithmic

   differentiation (take $\log$ of both sides, differentiate implicitly, multiply

   back by the original function) is the standard tool. Evaluating at $x=e$ is a

   deliberately friendly point since $\log e=1$ simplifies $\log(\log x)$ to $\log 1=0$.

   Step-by-Step Solution

   1. Term 1: $A=(\log x)^{1/x}$. Take logs: $\log A=\dfrac1x\log(\log x)$.
   2. Differentiate: $\dfrac{A'}{A}=-\dfrac{1}{x^2}\log(\log x)+\dfrac1x\cdot\dfrac{1}{x\log x} =-\dfrac{\log(\log x)}{x^2}+\dfrac{1}{x^2\log x}$.
   3. At $x=e$: $\log x=1\Rightarrow\log(\log x)=\ln1=0$, so $\dfrac{A'}{A}=\dfrac{1}{e^2\cdot1}=\dfrac1{e^2}$. Also $A=(\log e)^{1/e}=1^{1/e}=1$. So $A'=\dfrac1{e^2}$.
   4. Term 2: $B=x^{\log x}$. Take logs: $\log B=\log x\cdot\log x=(\log x)^2$. …

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8. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If $f(x) = x^{\sec^{-1}x}$, then $f'(2) =$ (A) $\dfrac{2^{\pi/3}}{6}\left(\pi - \sqrt{3}\log 2\right)$ (B) $\dfrac{2^{\pi/6}}{6}\left(\pi + \sqrt{3}\log 2\right)$ (C) $\dfrac{2^{\pi/3}}{6}\left(\pi + \sqrt{3}\log 2\right)$ (D) $\dfrac{2^{\pi/6}}{6}\left(\pi - \sqrt{3}\log 2\right)$

   ›Reveal solutionSolution

   Logarithmic differentiation of $x^{\sec^{-1}x}$, evaluated using $\sec^{-1}2=\pi/3$. Answer: $f'(2)=\dfrac{2^{\pi/3}}{6}(\pi+\sqrt3\log2)$.

   Concept and Intuition

   Whenever both the base and the exponent are functions of $x$ (here $x$ raised to $\sec^{-1}x$), the cleanest route is logarithmic differentiation: write $f=e^{g(x)\log x}$ and differentiate the exponent.

   Step-by-Step Solution

   1. $f(x)=x^{\sec^{-1}x} = e^{\sec^{-1}x\cdot\log x}$.
   2. $f'(x) = f(x)\cdot\dfrac{d}{dx}\left[\sec^{-1}x\cdot\log x\right] = f(x)\left[\left(\dfrac{1}{x\sqrt{x^2-1}}\right)\log x + \sec^{-1}x\cdot\dfrac1x\right]$ (using $\dfrac{d}{dx}\sec^{-1}x=\dfrac{1}{x\sqrt{x^2-1}}$ for $x>1$).
   3. At $x=2$: $\sec^{-1}(2)=\dfrac{\pi}{3}$ (since $\sec(\pi/3)=2$), and $\sqrt{x^2-1}=\sqrt3$.
   4. $f(2)=2^{\pi/3}$.
   5. Bracket at $x=2$: $\dfrac{\ln2}{2\sqrt3}+\dfrac{\pi/3}{2} = \dfrac{\ln2}{2\sqrt3}+\dfrac{\pi}{6}$.
   6. Put over denominator $6$: $\dfrac{\ln2}{2\sqrt3}=\dfrac{\sqrt3\ln2}{6}$ (rationalising), so bracket $=\dfrac{\pi+\sqrt3\ln2}{6}$. …

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9. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.$f(x)$ is a continuous function on $\mathbb{R}$ and $y = f(x)$ is a curve. If $(\alpha, \beta)$ is a point such that $\beta = f(\alpha)$ and $p\alpha + m\beta + n = 0$ ($p \ne 0, m \ne 0$), then which one of the following is True? (A) When $p + mf'(\alpha) = 0$, $px + my + n = 0$ intersects the curve $y = f(x)$ (B) $px + my + n = 0$ is always a tangent to the curve $y = f(x)$ (C) When $p + mf'(\alpha) \ne 0$, $px + my + n = 0$ intersects the curve $y = f(x)$ (D) $px + my + n = 0$ is never a tangent to the curve $y = f(x)$

   ›Reveal solutionSolution

   The line and curve always share $(\alpha,\beta)$; when the line's slope doesn't match the tangent slope there, the line crosses (genuinely intersects) the curve rather than merely touching it.

   Concept and Intuition

   $(\alpha,\beta)$ satisfies both $\beta=f(\alpha)$ (on the curve) and $p\alpha+m\beta+n=0$ (on the line), so the line and curve meet at this point regardless. The distinguishing question is whether the line is tangent there (touching without crossing, when slopes match) or genuinely crosses the curve (when slopes differ).

   Step-by-Step Solution

   1. The line $px+my+n=0$ has slope $-p/m$ (since $m\ne0$).
   2. The curve's tangent slope at $\alpha$ is $f'(\alpha)$.
   3. If $p+mf'(\alpha)=0$, i.e. $f'(\alpha)=-p/m$, the line's slope matches the tangent slope at $(\alpha,\beta)$ — the line behaves like the tangent there (touching, not necessarily crossing). …

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10. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.If $f(x) = \cot^{-1}\left(\frac{x^x + x^{-x}}{2}\right)$, then $f'(1) =$ (A) $1$ (B) $-1$ (C) $2$ (D) $-2$

    ›Reveal solutionSolution

    Recognizing $g(x)=\dfrac{x^x-x^{-x}}{2}=\sinh(x\log x)$ makes the derivative clean: $g(1)=0$, $g'(1)=1$, and the $\cot^{-1}$ chain rule gives $f'(1)=-1$.

    Concept and Intuition

    The combination $\dfrac{x^x - x^{-x}}{2}$ is exactly $\sinh(x\log x)$ (since $x^x=e^{x\log x}$), which vanishes at $x=1$ (because $x\log x=0$ there) — this structural fact makes the derivative of the outer $\cot^{-1}$ especially simple to evaluate at that point.

    Step-by-Step Solution

    1. Let $g(x)=\dfrac{x^x-x^{-x}}{2}$. Since $x^x=e^{x\log x}$ and $x^{-x}=e^{-x\log x}$, $g(x)=\sinh(x\log x)$.
    2. At $x=1$: $x\log x = 1\cdot0=0$, so $g(1)=\sinh(0)=0$.
    3. $g'(x) = \cosh(x\log x)\cdot\dfrac{d}{dx}(x\log x) = \cosh(x\log x)\,(\log x+1)$.
    4. At $x=1$: $\cosh(0)=1$ and $(\ln1+1)=1$, so $g'(1)=1\times1=1$. …

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11. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
    Q.The normal to the curve $y = f(x)$ at the points (3, 4) makes an angle $\dfrac{3\pi}{4}$ with positive x-axis then $f'(3) =$ (A) $3$ (B) $2$ (C) $1$ (D) $4$

    ›Reveal solutionSolution

    Convert the normal's angle to its slope, then use the perpendicularity of tangent and normal to find $f'(3)$. Answer: $f'(3)=1$.

    Concept and Intuition

    At any point on a curve, the tangent and the normal are perpendicular lines. If the normal makes angle $\theta$ with the positive x-axis, its slope is $\tan\theta$, and the tangent's slope is the negative reciprocal of that — which is exactly $f'(x)$ at that point, since the tangent's slope is by definition the derivative.

    Step-by-Step Solution

    1. Slope of normal $=\tan\left(\dfrac{3\pi}{4}\right)=\tan(135^\circ)=-1$.
    2. Tangent and normal are perpendicular, so (slope of tangent) $\times$ (slope of normal) $=-1$.
    3. Slope of tangent $=\dfrac{-1}{\text{slope of normal}}=\dfrac{-1}{-1}=1$. …

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12. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.$\displaystyle\int \cos^{-1}\left(\sqrt{\dfrac{x}{a+x}}\right) dx = f(x) + C \Rightarrow f'(a) =$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    $f'(x)$ is just the original integrand evaluated at $x$; plug in $x=a$ to get $\pi/4$.

    Concept and Intuition

    This question tests whether you realise you never need to actually perform the integration. If $\int g(x)\,dx = f(x)+C$, then by the very definition of an antiderivative $f'(x)=g(x)$ for every $x$ — so $f'(a)$ is simply the integrand evaluated at $x=a$.

    Step-by-Step Solution

    1. Here the integrand is $g(x)=\cos^{-1}\sqrt{\dfrac{x}{a+x}}$, and $f'(x)=g(x)$.
    2. So $f'(a)=\cos^{-1}\sqrt{\dfrac{a}{a+a}}=\cos^{-1}\sqrt{\dfrac{a}{2a}}=\cos^{-1}\sqrt{\dfrac12}$.
    3. $\sqrt{1/2}=1/\sqrt2=\cos(\pi/4)$, so $\cos^{-1}(1/\sqrt2)=\pi/4$. …

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