Q.Find $\frac{dy}{dx}$ in the following: $x^3 \log x$

Showing the 12 most recent of 48 on this concept.

1. 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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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $y = \begin{vmatrix} \sin x & \cos x & \sin x \\ \cos x & -\sin x & \cos x \\ x & 1 & 1 \end{vmatrix}$, then $\dfrac{dy}{dx} =$ (A) $x \cos x$ (B) $x \sin x$ (C) $\sin x + \cos x$ (D) $1$

   ›Reveal solutionSolution

   Expanding the determinant gives $y = x - 1$, so $\dfrac{dy}{dx} = 1$ — option (D).

   Working. Expand along the first row:

   $y = \sin x\,\big[(-\sin x)(1) - (\cos x)(1)\big] - \cos x\,\big[(\cos x)(1) - (\cos x)(x)\big] + \sin x\,\big[(\cos x)(1) - (-\sin x)(x)\big]$

   $= -\sin^2 x - \sin x\cos x \;-\; \cos^2 x\,(1-x) \;+\; \sin x\cos x + x\sin^2 x$

   The two $\sin x\cos x$ terms cancel, leaving: …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\lim_{x \to \frac{\pi}{3}} \dfrac{\tan^3 x - 3\tan x}{\cos\left(x + \dfrac{\pi}{6}\right)} =$ (A) $12$ (B) $24$ (C) $-24$ (D) $-12$

   ›Reveal solutionSolution

   A $0/0$ trig limit resolved by L'Hôpital's Rule; the answer is $-24$.

   Concept and Intuition

   Both numerator and denominator vanish at $x=\pi/3$ (since $\tan(\pi/3)=\sqrt3$ makes $\tan^3x-3\tan x=0$, and $x+\pi/6=\pi/2$ makes $\cos(x+\pi/6)=0$). This is a genuine $0/0$ form, so we differentiate numerator and denominator with respect to $x$ and substitute.

   Step-by-Step Solution

   1. Numerator $N(x)=\tan^3x - 3\tan x$. Its derivative: $N'(x) = 3\tan^2x\sec^2x - 3\sec^2x = 3\sec^2x(\tan^2x-1)$.
   2. At $x=\pi/3$: $\sec(\pi/3)=2 \Rightarrow \sec^2 = 4$; $\tan^2(\pi/3)=3 \Rightarrow \tan^2x-1=2$. So $N'(\pi/3)=3(4)(2)=24$.
   3. Denominator $D(x)=\cos(x+\pi/6)$. Its derivative: $D'(x)=-\sin(x+\pi/6)$.
   4. At $x=\pi/3$: $x+\pi/6=\pi/2$, and $\sin(\pi/2)=1$, so $D'(\pi/3)=-1$. …

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4. 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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5. 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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6. 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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7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If $\alpha$ and $\beta$ ($\alpha > \beta$) are the multiple roots of the equation $4x^4 + 4x^3 - 23x^2 - 12x + 36 = 0$, then $2\alpha - \beta =$ (A) $-1$ (B) $3$ (C) $5$ (D) $-7$

   ›Reveal solutionSolution

   The quartic factors neatly as $(x+2)^2(2x-3)^2$, so its two repeated ('multiple') roots are $-2$ and $\tfrac32$; with $\alpha>\beta$, $2\alpha-\beta=5$.

   Concept and Intuition

   'Multiple roots' of a polynomial are roots with multiplicity greater than $1$ (i.e., repeated roots). A quartic with two distinct double roots factors as $a(x-\alpha)^2(x-\beta)^2$ — recognizing this pattern is far faster than blindly applying the quartic formula.

   Step-by-Step Solution

   1. Try small integer/rational candidates in $4x^4+4x^3-23x^2-12x+36=0$. Testing $x=-2$: $4(16)+4(-8)-23(4)-12(-2)+36=64-32-92+24+36=0$. So $x=-2$ is a root.
   2. Synthetic division by $(x+2)$: $4x^4+4x^3-23x^2-12x+36 \div (x+2) = 4x^3-4x^2-15x+18$.
   3. Test $x=-2$ again in this cubic: $4(-8)-4(4)-15(-2)+18=-32-16+30+18=0$. So $x=-2$ is a repeated root (multiplicity at least 2).
   4. Divide again by $(x+2)$: $4x^3-4x^2-15x+18 \div (x+2) = 4x^2-12x+9$.
   5. Factor the quadratic: $4x^2-12x+9=(2x-3)^2$, giving the repeated root $x=\dfrac32$ (also multiplicity 2, since it's a perfect square).
   6. So the full factorization is $4x^4+4x^3-23x^2-12x+36=(x+2)^2(2x-3)^2$ (verified by re-expanding: $(x^2+4x+4)(4x^2-12x+9)=4x^4+4x^3-23x^2-12x+36$ ✓). …

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8. 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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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the tangent drawn at the point $(x_1, y_1)$, $x_1, y_1 \in N$ on the curve $y = x^4 - 2x^3 + x^2 + 5x$ passes through origin, then $x_1 + y_1 =$ (A) 5 (B) 4 (C) 7 (D) 6

   ›Reveal solutionSolution

   The tangent-through-origin condition gives an algebraic equation in $x_1$ that factors cleanly; the only natural-number root is $x_1=1$, giving $y_1=5$ and $x_1+y_1=6$.

   Concept and Intuition

   If the tangent at $(x_1,y_1)$ on a curve $y=f(x)$ passes through the origin, the slope of the tangent (which is $f'(x_1)$) must also equal the slope of the line from the origin to $(x_1,y_1)$, i.e. $\dfrac{y_1-0}{x_1-0}=\dfrac{y_1}{x_1}$. This gives $y_1=x_1f'(x_1)$, a clean algebraic condition combining the curve's value and its derivative at the same point.

   Step-by-Step Solution

   1. $f(x)=x^4-2x^3+x^2+5x$, so $y_1=f(x_1)=x_1^4-2x_1^3+x_1^2+5x_1$.
   2. $f'(x)=4x^3-6x^2+2x+5$, so the tangent slope at $x_1$ is $f'(x_1)=4x_1^3-6x_1^2+2x_1+5$.
   3. Tangent line through $(x_1,y_1)$ passing through the origin: slope $=\dfrac{y_1}{x_1}=f'(x_1)$ (for $x_1\ne0$), i.e. $y_1=x_1f'(x_1)$:

   $x_1^4-2x_1^3+x_1^2+5x_1 = x_1(4x_1^3-6x_1^2+2x_1+5) = 4x_1^4-6x_1^3+2x_1^2+5x_1$

   4. Simplify:

   $0 = (4x_1^4-6x_1^3+2x_1^2+5x_1) - (x_1^4-2x_1^3+x_1^2+5x_1) = 3x_1^4-4x_1^3+x_1^2$

   $= x_1^2(3x_1^2-4x_1+1) = x_1^2(3x_1-1)(x_1-1)$ …

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10. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.The slope of a tangent drawn at the point $P(\alpha, \beta)$ lying on the curve $y = \dfrac{1}{2x-5}$ is $-2$. If P lies in the fourth quadrant, then $\alpha - \beta =$ (A) 4 (B) 3 (C) 2 (D) 1

    ›Reveal solutionSolution

    Set the derivative equal to $-2$ to get two candidate $x$-values, then use the

    fourth-quadrant condition ($x>0,\ y<0$) to pick the right one: $\alpha-\beta=3$.

    Concept and Intuition

    The slope condition alone gives a quadratic (two candidate points on the curve),

    since $\frac{dy}{dx}=-2/(2x-5)^2$ is even in $(2x-5)$. The extra geometric

    condition — "P lies in the fourth quadrant" — is exactly what's needed to

    discard the spurious root and pin down a unique point.

    Step-by-Step Solution

    1. $y=\dfrac{1}{2x-5}=(2x-5)^{-1}$, so $\dfrac{dy}{dx}=-2(2x-5)^{-2}=\dfrac{-2}{(2x-5)^2}$.
    2. Set slope $=-2$: $\dfrac{-2}{(2x-5)^2}=-2\ \Rightarrow\ (2x-5)^2=1\ \Rightarrow\ 2x-5=\pm1$.
    3. $2x-5=1\Rightarrow x=3$; $2x-5=-1\Rightarrow x=2$.
    4. At $x=3$: $y=\dfrac1{2(3)-5}=\dfrac11=1$. Point $(3,1)$ — first quadrant ($x>0,y>0$), rejected. …

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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 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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