Q.If $x = e^{\cos 2t}$ and $y = e^{\sin 2t}$, prove that $\dfrac{dy}{dx} = -\dfrac{y \log x}{x \log y}$.

Showing the 12 most recent of 50 on this concept.

1. 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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2. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $y^{y^{y^{\cdot^{\cdot^{\cdot^{\infty}}}}}} = \log\{x + \log\{x + \cdots\}\}$, then $\dfrac{dy}{dx}$ at $x = e^2 - 2$, $y = \sqrt{2}$ equals _____ (A) $\dfrac{\log 2}{2\sqrt{2}(e^2 - 1)}$ (B) $\dfrac{1 - \log 2}{2\sqrt{2}(e^2 - 1)}$ (C) $\dfrac{\sqrt{2}(1 - \log 2)}{e^2 - 1}$ (D) $\dfrac{\log 2}{\sqrt{2}(e^2 - 1)}$

   ›Reveal solutionSolution

   Both sides define the same implicit quantity $u$ via a self-referential equation; differentiate each side's defining equation implicitly and combine using the chain rule. The answer is (B).

   Concept and Intuition

   The infinite power tower $y^{y^{y^{\cdots}}}=u$ satisfies the self-consistency equation $u=y^u$ (the tower "regenerates" itself). Likewise the infinite nested logarithm $\log\{x+\log\{x+\cdots\}\}=u$ satisfies $u=\log(x+u)$. Since the problem states these two quantities are equal (both equal to the same $u$), $u$ is implicitly a common function linking $x$ and $y$; differentiating each defining relation gives $du/dy$ and $du/dx$, and the chain rule combines them into $dy/dx$.

   Step-by-Step Solution

   1. Verify $u=2$ at the given point. Nested log: $u=\log(x+u)$ at $x=e^2-2$: try $u=2$: $\log(e^2-2+2)=\log(e^2)=2$ ✓. Tower: $u=y^u$ at $y=\sqrt2$: try $u=2$: $(\sqrt2)^2=2$ ✓. Both consistent with $u=2$.
   2. Differentiate the tower relation $u=y^u$ w.r.t. $y$. Take $\log$: $\log u = u\log y$. Differentiate: $\dfrac1u\dfrac{du}{dy} = \dfrac{du}{dy}\log y + \dfrac uy$ $\Rightarrow \left(\dfrac1u-\log y\right)\dfrac{du}{dy}=\dfrac uy \Rightarrow \dfrac{du}{dy}=\dfrac{u^2}{y(1-u\log y)}$. At $y=\sqrt2,\ u=2$: $\dfrac{du}{dy}=\dfrac{4}{\sqrt2(1-2\log\sqrt2)}=\dfrac{4}{\sqrt2(1-\ln2)}=\dfrac{2\sqrt2}{1-\ln2}$. …

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3. 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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4. 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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5. 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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6. 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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7. 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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8. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $(a + \sqrt{2}\,b\cos x)(a - \sqrt{2}\,b\cos y) = a^2 - b^2$ where $a > b > 0$, then at $\left(\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)$, $\dfrac{dy}{dx} =$ (A) $\dfrac{a + b}{a - b}$ (B) $\dfrac{a - b}{a + b}$ (C) $\dfrac{a - 2b}{a + 2b}$ (D) $\dfrac{2a + b}{2a - b}$

   ›Reveal solutionSolution

   Expand the product, simplify by dividing by the common factor $b$, then implicitly differentiate and evaluate at $x=y=\pi/4$. Answer: $\dfrac{a-b}{a+b}$.

   Concept and Intuition

   The given relation looks intimidating as a product, but expanding it cancels the $a^2$ on both sides (since the RHS is $a^2-b^2$) and leaves a much simpler equation relating $\cos x,\cos y$, and $\cos x\cos y$. From there it's routine implicit differentiation; the special evaluation point $x=y=\pi/4$ is chosen because $\sin$ and $\cos$ coincide there, which cancels neatly.

   Step-by-Step Solution

   1. Expand: $a^2 - a\sqrt2 b\cos y + a\sqrt2 b\cos x - 2b^2\cos x\cos y = a^2-b^2$.
   2. Cancel $a^2$ from both sides: $\sqrt2 ab(\cos x-\cos y) - 2b^2\cos x\cos y = -b^2$.
   3. Divide through by $b$ (nonzero): $\sqrt2 a(\cos x-\cos y) - 2b\cos x\cos y + b = 0$.
   4. Differentiate implicitly w.r.t. $x$ (treat $y=y(x)$):

   $\sqrt2 a\left(-\sin x + \sin y\cdot y'\right) - 2b\left(-\sin x\cos y - \cos x\sin y\cdot y'\right) = 0.$

   5. Group $y'$ terms: $y'\left(\sqrt2 a\sin y + 2b\cos x\sin y\right) = \sqrt2 a\sin x - 2b\sin x\cos y$.
   6. So $y' = \dfrac{\sin x(\sqrt2 a - 2b\cos y)}{\sin y(\sqrt2 a + 2b\cos x)}$.
   7. At $x=y=\dfrac\pi4$: $\sin x=\sin y=\cos x=\cos y=\dfrac1{\sqrt2}$. Substitute: …

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9. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\sin x \sqrt{\cos y} - \cos y \sqrt{\sin x} = 0$, then $\dfrac{dy}{dx} =$ (A) $\tan x$ (B) $1$ (C) $-1$ (D) $-\cot x$

   ›Reveal solutionSolution

   The given relation simplifies to $\sin x=\cos y$; implicit differentiation of this simpler relation gives $\dfrac{dy}{dx}=-1$.

   Concept and Intuition

   Many implicit-differentiation problems hide a much simpler relation inside a more complicated-looking equation. Recognizing that both sides share a common factor of $\sqrt{\sin x\,\cos y}$ lets us cancel down to something we can differentiate directly, instead of differentiating the square-root expression term by term.

   Step-by-Step Solution

   1. Start with $\sin x\sqrt{\cos y}-\cos y\sqrt{\sin x}=0$, i.e. $\sin x\sqrt{\cos y}=\cos y\sqrt{\sin x}$.
   2. Divide both sides by $\sqrt{\sin x}\,\sqrt{\cos y}$ (both taken positive for the relevant domain):

   $\frac{\sin x}{\sqrt{\sin x}} = \frac{\cos y}{\sqrt{\cos y}} \;\Rightarrow\; \sqrt{\sin x}=\sqrt{\cos y}.$

   3. Squaring, $\sin x=\cos y$.
   4. Differentiate both sides with respect to $x$: $\cos x = -\sin y\,\dfrac{dy}{dx}$. …

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10. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.If $\mathrm{Tan}^{-1}x^2 + \mathrm{Tan}^{-1}y^2 = \dfrac{\pi}{2}$, then $\left(\dfrac{dy}{dx}\right)_{(-1,2)} =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $-\dfrac{1}{2}$

    ›Reveal solutionSolution

    Reducing to $y^2=x^{-2}$ gives $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$, which at $(-1,2)$ equals $\dfrac12$.

    Concept and Intuition

    If $\mathrm{Tan}^{-1}a+\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}$ with $a,b>0$, then $\mathrm{Tan}^{-1}b=\dfrac{\pi}{2}-\mathrm{Tan}^{-1}a=\mathrm{Cot}^{-1}a$, so $b=\dfrac{1}{a}$. Applying this to $a=x^2,\ b=y^2$ collapses the relation into an algebraic one.

    Step-by-Step Solution

    1. From $\mathrm{Tan}^{-1}x^2+\mathrm{Tan}^{-1}y^2=\dfrac{\pi}{2}$ we get $y^2=\dfrac{1}{x^2}=x^{-2}$.
    2. Differentiate: $2y\,\dfrac{dy}{dx}=-2x^{-3}$.
    3. Hence $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$. …

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11. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.m is the slope of a tangent to the curve $e^y = 1 + x^2$ at $x = 1$ then $m =$ (A) $\dfrac{2}{\log 2}$ (B) $\log 2$ (C) $2$ (D) $1$

    ›Reveal solutionSolution

    Implicit differentiation of $e^y=1+x^2$ gives slope $2x/(1+x^2)$, which is $1$ at $x=1$.

    Concept and Intuition

    This is a straightforward implicit differentiation: differentiate both sides with respect to $x$, treating $y$ as a function of $x$, then substitute the known relation back in to eliminate $e^y$.

    Step-by-Step Solution

    1. Differentiate $e^y = 1+x^2$ with respect to $x$: $e^y\dfrac{dy}{dx} = 2x$.
    2. So $\dfrac{dy}{dx} = \dfrac{2x}{e^y} = \dfrac{2x}{1+x^2}$ (substituting $e^y=1+x^2$ from the original equation). …

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12. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.If $\log\left(\sqrt{1+x^2}-x\right)=y\left(\sqrt{1+x^2}\right)$, then $(1+x^2)\dfrac{dy}{dx}+xy=$ (A) $0$ (B) $1$ (C) $2$ (D) $-1$

    ›Reveal solutionSolution

    Implicit differentiation of $\log(\sqrt{1+x^2}-x)=y\sqrt{1+x^2}$, using $(\sqrt{1+x^2}-x)(\sqrt{1+x^2}+x)=1$, collapses directly to the requested combination. Answer: $-1$.

    Concept and Intuition

    Writing $s=\sqrt{1+x^2}$ turns the relation into $\log(s-x)=ys$, a compact form whose derivative — after using the identity $s^2-x^2=1$ — telescopes into exactly the expression $(1+x^2)y'+xy$ asked for.

    Step-by-Step Solution

    1. Let $s=\sqrt{1+x^2}$; then $s'=\dfrac{x}{s}$ and $s^2-x^2=1\Rightarrow(s-x)(s+x)=1\Rightarrow\dfrac{1}{s-x}=s+x$.
    2. Given: $\log(s-x)=ys$. Differentiate both sides w.r.t. $x$: $\frac{s'-1}{s-x}=y's+ys'$ …

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