Q.Find $\frac{dy}{dx}$ in the following: $x^{x-3} + (x-3)^x$, for $x > 3$

Showing the 12 most recent of 50 on this concept.

1. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $y = (\log_x \sin x)^x$, then $\dfrac{dy}{dx} =$ (A) $y\left[\dfrac{x\sin x}{\log\cos x} + \log(\log\sin x) + \dfrac{1}{\log x} - \log(\log x)\right]$ (B) $y\left[\dfrac{x\cos x}{\log\sin x} - \log(\log\sin x) + \dfrac{1}{\log x} + \log(\log x)\right]$ (C) $y\left[\dfrac{x\cot x}{\log\sin x} + \log(\log\sin x) - \dfrac{1}{\log x} - \log(\log x)\right]$ (D) $y\left[\dfrac{x\cot x}{\log\sin x} - \log(\log\sin x) + \dfrac{1}{\log x} - \log x\right]$

   ›Reveal solutionSolution

   This is a logarithmic-differentiation problem with a function-of-a-function base; careful chain-rule bookkeeping on $u=\log_x(\sin x)$ gives option (C).

   Concept and Intuition

   When both the base and the exponent are functions of $x$ (here the base is itself $\log_x(\sin x)$), the standard technique is logarithmic differentiation: take $\ln$ of both sides to turn the power into a product, then differentiate using the product and chain rules.

   Step-by-Step Solution

   1. Let $u = \log_x(\sin x) = \dfrac{\ln \sin x}{\ln x}$, so $y = u^x$.
   2. Take logs: $\ln y = x\ln u$.
   3. Differentiate: $\dfrac{y'}{y} = \ln u + x\cdot\dfrac{u'}{u}$.
   4. Compute $u'$: with $u=\dfrac{\ln\sin x}{\ln x}$,

   $u' = \frac{\cot x\cdot \ln x - \ln\sin x\cdot \frac1x}{(\ln x)^2}.$

   5. Then

   $\frac{u'}{u} = \frac{\cot x\cdot\ln x - \frac{\ln\sin x}{x}}{(\ln x)^2}\cdot\frac{\ln x}{\ln\sin x} = \frac{\cot x}{\ln\sin x} - \frac{1}{x\ln x}.$

   6. So $x\cdot\dfrac{u'}{u} = \dfrac{x\cot x}{\ln\sin x} - \dfrac{1}{\ln x}$.
   7. And $\ln u = \ln(\ln\sin x) - \ln(\ln x)$.
   8. Combine:

   $\frac{y'}{y} = \ln(\ln\sin x) - \ln(\ln x) + \frac{x\cot x}{\ln\sin x} - \frac{1}{\ln x},$

   so …

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2. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $x^x y^y = e^e$, then $\left(\dfrac{d^2y}{dx^2}\right)_{(e,e)} =$ (A) $\dfrac{1}{e}\left(\dfrac{dy}{dx}\right)_{(e,e)}$ (B) $\left(\dfrac{dy}{dx}\right)_{(e,e)} + \dfrac{1}{e}$ (C) $\left(\dfrac{dy}{dx}\right)_{(e,e)} - \dfrac{1}{e}$ (D) $e\left(\dfrac{dy}{dx}\right)_{(e,e)}$

   ›Reveal solutionSolution

   Logarithmic differentiation of $x^xy^y=e^e$ twice, evaluated at $(e,e)$, shows the second derivative equals $\frac1e$ times the first derivative there.

   Concept and Intuition

   Expressions like $x^x$ are best handled by taking logs first (since $\log(x^x)=x\log x$ is much easier to differentiate than $x^x$ directly). Implicit differentiation then relates $y'$ and $y''$ through the resulting equation.

   Step-by-Step Solution

   1. Take log: $x\log x + y\log y = \log(e^e) = e$ (a constant).
   2. Differentiate w.r.t. $x$: $(\log x+1) + (\log y+1)y' = 0$.
   3. Solve: $y' = -\dfrac{\log x+1}{\log y+1}$. At $(e,e)$: $\log e=1$, so $y'=-\dfrac{2}{2}=-1$.
   4. Differentiate the relation $(\log x+1)+(\log y+1)y'=0$ again w.r.t. $x$: $\dfrac1x + \dfrac{(y')^2}{y} + (\log y+1)y'' = 0$. …

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3. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If $y=(\tan x)^{\sin x}$, then $\dfrac{dy}{dx}=$ (A) $(\tan x)^{\sin x}\{\sec x+(\cos x)(\log(\tan x))\}$ (B) $(\sin x)^{\tan x}\{\sec x+(\cos x)(\log(\tan x))\}$ (C) $(\tan x)^{\sin x}\{\sec x-(\cos x)(\log(\tan x))\}$ (D) $(\sin x)^{\tan x}\{\sec x-(\cos x)(\log(\tan x))\}$

   ›Reveal solutionSolution

   Logarithmic differentiation of $y=(\tan x)^{\sin x}$ gives $y' = (\tan x)^{\sin x}\{\sec x+\cos x\log(\tan x)\}$.

   Concept and Intuition

   Whenever both the base and the exponent are functions of $x$ (here base $\tan x$, exponent $\sin x$), take natural log of both sides first — this converts the power into a product, which is easy to differentiate using the product rule.

   Step-by-Step Solution

   1. $y=(\tan x)^{\sin x}$. Take $\log$: $\log y = \sin x \cdot \log(\tan x)$.
   2. Differentiate both sides w.r.t. $x$ using the product rule on the right: $\dfrac{1}{y}\dfrac{dy}{dx} = \cos x \cdot \log(\tan x) + \sin x \cdot \dfrac{1}{\tan x}\cdot \sec^2 x$.
   3. Simplify the second term: $\sin x \cdot \dfrac{\sec^2 x}{\tan x} = \sin x \cdot \dfrac{1}{\cos^2 x}\cdot\dfrac{\cos x}{\sin x} = \dfrac{1}{\cos x} = \sec x$.
   4. So $\dfrac1y \dfrac{dy}{dx} = \cos x\,\log(\tan x) + \sec x$. …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $x^2y - xy^2 + x^3 - y^3 = 0$, then $\dfrac{dy}{dx}$ at the point $(1, 1)$ is (A) $1$ (B) $0$ (C) $-1$ (D) Does not exist

   ›Reveal solutionSolution

   Implicit differentiation of a symmetric cubic curve, evaluated at the point $(1,1)$.

   Concept and Intuition

   When a curve is given implicitly (not solved for $y$), differentiate every term with respect to $x$, treating $y$ as a function of $x$ and applying the product rule wherever $x$ and $y$ appear together. Collecting all the $y'$ terms on one side isolates the slope as a ratio of two expressions in $x,y$.

   Step-by-Step Solution

   1. Differentiate term by term: $\dfrac{d}{dx}(x^2y)=2xy+x^2y'$; $\dfrac{d}{dx}(xy^2)=y^2+2xyy'$; $\dfrac{d}{dx}(x^3)=3x^2$; $\dfrac{d}{dx}(y^3)=3y^2y'$.
   2. So $2xy+x^2y'-y^2-2xyy'+3x^2-3y^2y'=0$.
   3. Collect $y'$ terms: $y'(x^2-2xy-3y^2)=-(2xy-y^2+3x^2)$, i.e. $y'=\dfrac{-(2xy-y^2+3x^2)}{x^2-2xy-3y^2}=\dfrac{y^2-2xy-3x^2}{x^2-2xy-3y^2}$. …

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5. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $\tan(e^{3x}) = \cot(e^{2y})$, then at $x = 0$, $\dfrac{dy}{dx} =$ (A) $\dfrac{3}{2-\pi}$ (B) $\dfrac{2-\pi}{3}$ (C) $\dfrac{3}{\pi-2}$ (D) $\dfrac{\pi-2}{3}$

   ›Reveal solutionSolution

   Rewrite $\cot$ as a shifted $\tan$ to turn the equation into an algebraic (exponential) relation between $x$ and $y$, then implicitly differentiate and evaluate at $x=0$. Answer: $\dfrac{3}{2-\pi}$.

   Concept and Intuition

   $\tan\theta_1=\tan\theta_2$ implies $\theta_1=\theta_2+n\pi$ for integer $n$; taking $n=0$ (the principal relation intended here) converts the trig equation into a clean equation between the exponential expressions, which we can differentiate implicitly.

   Step-by-Step Solution

   1. Use the identity $\cot\theta = \tan\left(\dfrac{\pi}{2}-\theta\right)$ with $\theta=e^{2y}$: $\cot(e^{2y}) = \tan\left(\dfrac{\pi}{2}-e^{2y}\right)$.
   2. Given $\tan(e^{3x}) = \cot(e^{2y}) = \tan\left(\dfrac{\pi}{2}-e^{2y}\right)$, equate arguments (principal branch): $e^{3x} = \dfrac{\pi}{2}-e^{2y}$.
   3. Rearrange: $e^{3x}+e^{2y} = \dfrac{\pi}{2}$.
   4. Differentiate both sides w.r.t. $x$: $3e^{3x} + 2e^{2y}\dfrac{dy}{dx} = 0$.
   5. Solve: $\dfrac{dy}{dx} = -\dfrac{3e^{3x}}{2e^{2y}}$.
   6. At $x=0$: $e^{3x}=e^0=1$. From step 3, $1+e^{2y}=\dfrac{\pi}{2}\Rightarrow e^{2y}=\dfrac{\pi}{2}-1=\dfrac{\pi-2}{2}$. …

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6. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $\dfrac{y}{x}\cos^4\alpha + \dfrac{x}{y}\sin^4\alpha = 2\sin^2\alpha \cdot \cos^2\alpha$, then $\dfrac{dy}{dx} =$ (A) $\sin^3\alpha \cos\alpha$ (B) $\sin^2\alpha \cos^2\alpha$ (C) $\dfrac{\sin^2\alpha}{\cos^2\alpha}$ (D) $\sin\alpha \cos^3\alpha$

   ›Reveal solutionSolution

   The given relation is secretly a perfect square in disguise; it forces $y=x\tan^2\alpha$, so $dy/dx=\tan^2\alpha$.

   Concept and Intuition

   Rather than differentiating implicitly right away, it pays to recognise the algebraic structure first. Multiplying by $xy$ converts the equation into a quadratic in $x$ and $y$ that factors as a perfect square, revealing $y/x$ is actually a constant (independent of $x$), which makes the derivative trivial.

   Step-by-Step Solution

   1. Start from $\dfrac{y}{x}\cos^4\alpha+\dfrac{x}{y}\sin^4\alpha = 2\sin^2\alpha\cos^2\alpha$.
   2. Multiply both sides by $xy$: $y^2\cos^4\alpha + x^2\sin^4\alpha = 2xy\sin^2\alpha\cos^2\alpha$.
   3. Rearrange: $y^2\cos^4\alpha - 2xy\sin^2\alpha\cos^2\alpha + x^2\sin^4\alpha = 0$.
   4. This is $(y\cos^2\alpha - x\sin^2\alpha)^2 = 0$, so $y\cos^2\alpha = x\sin^2\alpha$, i.e. $y = x\tan^2\alpha$. …

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7. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.If $y = \sqrt{x + \sqrt{x + \sqrt{x + \cdots \infty}}}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{1}{y}$ (B) $\dfrac{1}{x}$ (C) $\dfrac{1}{2x-1}$ (D) $\dfrac{1}{2y-1}$

   ›Reveal solutionSolution

   The infinite nested radical satisfies $y^2=x+y$ (self-similarity), which is then differentiated implicitly.

   Concept and Intuition

   An infinitely repeating nested expression under a radical satisfies a self-referential equation: the whole expression $y$ equals the same structure with $x+y$ under the first radical (since removing the outermost layer just reproduces $y$ again).

   Step-by-Step Solution

   1. $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=\sqrt{x+y}$ (the inner infinite tail is again $y$).
   2. Square both sides: $y^2=x+y$.
   3. Differentiate implicitly with respect to $x$: $2y\dfrac{dy}{dx}=1+\dfrac{dy}{dx}$.
   4. Collect: $\dfrac{dy}{dx}(2y-1)=1\Rightarrow\dfrac{dy}{dx}=\dfrac1{2y-1}$. …

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8. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $x^2+y^2=t+\dfrac{1}{t}$ and $x^4+y^4=t^2+\dfrac{1}{t^2}$, then $x^3 y\dfrac{dy}{dx}=$ (A) -1 (B) 0 (C) 1 (D) 2

   ›Reveal solutionSolution

   The two given relations force $x^2y^2=1$ (i.e. $xy$ is constant), from which $x^3y\,dy/dx=-1$.

   Concept and Intuition

   Rather than solving for $x,y$ in terms of $t$ explicitly, combine the two given equations algebraically (square the first, subtract the second) to eliminate $t$ entirely and land on a simple constant-product relation between $x$ and $y$.

   Step-by-Step Solution

   1. Square the first relation: $(x^2+y^2)^2=\left(t+\dfrac1t\right)^2=t^2+2+\dfrac{1}{t^2}$, i.e.

   $x^4+2x^2y^2+y^4=t^2+2+\frac{1}{t^2}$

   2. The second given relation is $x^4+y^4=t^2+\dfrac{1}{t^2}$.
   3. Subtract: $2x^2y^2 = \left(t^2+2+\dfrac1{t^2}\right)-\left(t^2+\dfrac1{t^2}\right)=2$, so $x^2y^2=1$.
   4. This means $xy=\pm1$, a constant independent of $t$. Differentiate $xy=\text{const}$ implicitly: …

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9. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If the locus of the points on the curve $x^3y^2+\dfrac{x^2}{y}=5$ at which the tangent is parallel to X-axis is $f(x,y)=0$, then the point that lies on this curve $f(x,y)=0$ is (A) $(2,\sqrt[3]{3})$ (B) $(\sqrt[3]{2},3)$ (C) $\left(-2,\dfrac{1}{\sqrt[3]{3}}\right)$ (D) $\left(-\sqrt[3]{2},\dfrac{1}{\sqrt[3]{3}}\right)$

   ›Reveal solutionSolution

   Setting $dy/dx=0$ in the implicit differentiation of the curve gives the locus condition $3xy^3+2=0$; testing the four points, only $\left(-2,3^{-1/3}\right)$ satisfies it.

   Concept and Intuition

   "Tangent parallel to the X-axis" means $dy/dx=0$ at that point. Differentiating the curve implicitly and substituting $y'=0$ eliminates the derivative, leaving a plain algebraic relation between $x$ and $y$ — this relation is the locus equation $f(x,y)=0$, and we just need to check which candidate point satisfies it.

   Step-by-Step Solution

   1. Curve: $x^3y^2+x^2y^{-1}=5$.
   2. Differentiate w.r.t. $x$: $3x^2y^2+x^3\cdot2yy'+2xy^{-1}+x^2(-y^{-2})y'=0$, i.e. $3x^2y^2+2x^3yy'+\dfrac{2x}{y}-\dfrac{x^2y'}{y^2}=0$.
   3. Set $y'=0$ (horizontal tangent): $3x^2y^2+\dfrac{2x}{y}=0$.
   4. Multiply through by $y$: $3x^2y^3+2x=0 \Rightarrow x(3xy^3+2)=0$. Since $x=0$ doesn't satisfy the original curve, $x\ne0$, so $3xy^3+2=0 \Rightarrow xy^3=-\dfrac23$.
   5. Test each option against $xy^3=-\dfrac23$:
      • (A) $(2,3^{1/3})$: $xy^3=2\cdot3=6$ ✗
      • (B) $(2^{1/3},3)$: $xy^3=2^{1/3}\cdot27$ ✗ …

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10. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.If $y = \log_y x$, then $\dfrac{dy}{dx} =$ ______ (A) $\dfrac{1}{x\log y}$ (B) $\dfrac{\log y}{x(1+\log y)}$ (C) $\dfrac{1}{x(1+\log y)}$ (D) $\dfrac{1}{1+\log y}$

    ›Reveal solutionSolution

    Rewriting $y=\log_y x$ as $y\ln y=\ln x$ and differentiating implicitly gives $\dfrac{dy}{dx}=\dfrac{1}{x(1+\log y)}$.

    Concept and Intuition

    $\log_y x$ means "logarithm of $x$ to base $y$", i.e. $\dfrac{\ln x}{\ln y}$. So the given relation $y=\log_y x$ really means $y=\dfrac{\ln x}{\ln y}$, or equivalently $y\ln y=\ln x$ — a cleaner form to differentiate implicitly, since it avoids a quotient with $y$ in both places.

    Step-by-Step Solution

    1. $y=\log_y x=\dfrac{\ln x}{\ln y} \Rightarrow y\ln y=\ln x$.
    2. Differentiate both sides with respect to $x$, treating $y$ as a function of $x$:

    $\frac{d}{dx}(y\ln y)=\frac{d}{dx}(\ln x)$

    3. LHS (product rule): $\dfrac{dy}{dx}\ln y+y\cdot\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{dy}{dx}(\ln y+1)$.
    4. RHS: $\dfrac1x$.
    5. So $\dfrac{dy}{dx}(\ln y+1)=\dfrac1x \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x(1+\ln y)}=\dfrac{1}{x(1+\log y)}$. …

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11. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.If $x^3 - 2x^2y^2 + 5x + y - 5 = 0$, then at $(1,1)$, $y''(1) =$ (A) $-\frac{197}{27}$ (B) $\frac{125}{31}$ (C) $12$ (D) $-\frac{238}{27}$

    ›Reveal solutionSolution

    Implicit differentiation twice, using the values at $(1,1)$ and the first-derivative value $y'(1)=4/3$, gives $y''(1) = -238/27$.

    Concept and Intuition

    For a curve defined implicitly by $F(x,y)=0$, we differentiate throughout with respect to $x$, treating $y$ as a function of $x$ (chain rule at every $y$-term), to get an equation involving $y'$; solving that gives $y'$ in terms of $x,y$. Differentiating that resulting equation once more (again using the chain rule, now also needing $y'$ itself) and substituting known values gives $y''$.

    Step-by-Step Solution

    1. Verify $(1,1)$ lies on the curve: $1-2+5+1-5=0$. ✓
    2. Differentiate $x^3-2x^2y^2+5x+y-5=0$ w.r.t. $x$: $3x^2 - 2(2xy^2+2x^2yy') + 5+y' = 0 \Rightarrow 3x^2-4xy^2-4x^2yy'+5+y'=0$.
    3. Collect $y'$ terms: $y'(1-4x^2y) = -3x^2+4xy^2-5$.
    4. At $(1,1)$: numerator $=-3(1)+4(1)(1)-5=-4$; denominator $=1-4(1)(1)=-3$. So $y'(1)=\dfrac{-4}{-3}=\dfrac43$.
    5. Differentiate $y'(1-4x^2y)=-3x^2+4xy^2-5$ again w.r.t. $x$ (product rule on the LHS, chain rule throughout): LHS derivative: $y''(1-4x^2y) + y'\cdot\big(-(8xy+4x^2y')\big)$. RHS derivative: $-6x+4(y^2+2xyy')$. …

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12. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If $x^2 + y^2 = t - \dfrac{1}{t}$ and $x^4 + y^4 = t^2 + \dfrac{1}{t^2}$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{y}{x}$ (B) $\dfrac{y^2}{x^2}$ (C) $\sqrt{\dfrac{y}{x}}$ (D) $-\dfrac{y}{x}$

    ›Reveal solutionSolution

    Eliminating the parameter $t$ between the two given equations produces the direct relation $x^2y^2=-1$ between $x$ and $y$, whose implicit derivative is $-y/x$.

    Concept and Intuition

    When $x$ and $y$ are both linked to a parameter $t$ through two equations, differentiating each with respect to $t$ separately (and dividing) works, but it is often faster — and here it is exact — to first eliminate $t$ algebraically to get a direct $x$–$y$ relation, then differentiate that implicitly in the ordinary way.

    Step-by-Step Solution

    1. Square the first equation: $(x^2+y^2)^2=\left(t-\dfrac1t\right)^2=t^2-2+\dfrac1{t^2}$, i.e. $x^4+2x^2y^2+y^4=t^2+\dfrac1{t^2}-2$.
    2. The second equation says $x^4+y^4=t^2+\dfrac1{t^2}$. Substitute this in: $\left(t^2+\dfrac1{t^2}\right)+2x^2y^2=t^2+\dfrac1{t^2}-2$.
    3. This forces $2x^2y^2=-2\Rightarrow x^2y^2=-1$ — a $t$-free relation directly linking $x$ and $y$. …

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