Q.Find $\dfrac{dy}{dx}$ when $x$ and $y$ are connected by the relation: $\sec(x + y) = xy$.

Showing the 12 most recent of 50 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $3\sin xy + 4\cos xy = 5$, then $\dfrac{dy}{dx}$ is equal to ____ (A) $\dfrac{3\sin xy + 4\cos xy}{3\cos xy - 4\sin xy}$ (B) $\dfrac{3\cos xy + 4\sin xy}{4\cos xy - 3\sin xy}$ (C) $\dfrac{-y}{x}$ (D) $\dfrac{x}{y}$

   ›Reveal solutionSolution

   Since $3\sin\theta+4\cos\theta$ has maximum value exactly $5$ (as $\sqrt{3^2+4^2}=5$), equating it to $5$ forces $xy$ to be a fixed constant, so implicit differentiation of $xy=c$ gives $dy/dx=-y/x$.

   Concept and Intuition

   $a\sin\theta+b\cos\theta$ always has amplitude $\sqrt{a^2+b^2}$ — here $\sqrt{9+16}=5$. So the equation $3\sin(xy)+4\cos(xy)=5$ isn't a "generic" implicit curve; it can only be satisfied when the expression sits exactly at its maximum, which happens at one specific angle. That pins $xy$ to a single constant value, turning a trigonometric-looking implicit relation into the much simpler $xy=\text{const}$.

   Step-by-Step Solution

   1. Note $3^2+4^2=25=5^2$, so $3\sin\theta+4\cos\theta$ has maximum value $5$, attained only when $\theta$ equals the specific angle $\phi=\tan^{-1}(3/4)$ (mod $2\pi$).
   2. The given equation demands $3\sin(xy)+4\cos(xy)=5$, i.e. the maximum — so $xy=\phi$ is fixed, a constant independent of which point on the curve we pick. …

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2. 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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3. 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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4. 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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5. 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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6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\sqrt{x-xy}+\sqrt{y-xy}=1$, then $\dfrac{dy}{dx}=$ (A) $-\sqrt{\dfrac{y-y^2}{x-x^2}}$ (B) $-\sqrt{\dfrac{1-y^2}{1-x^2}}$ (C) $-\sqrt{\dfrac{1-y}{1-x}}$ (D) $-\sqrt{\dfrac{x-y}{x+y}}$

   ›Reveal solutionSolution

   Squaring the constraint reveals that it forces $x+y=1$ identically, so $\dfrac{dy}{dx}=-1$ throughout; matching this against the options singles out $-\sqrt{\dfrac{y-y^2}{x-x^2}}$, since it equals $-1$ for every point satisfying $y=1-x$.

   Concept and Intuition

   Rather than blindly grinding through implicit differentiation of two square roots, it pays to first understand the curve itself. Squaring $\sqrt{x-xy}+\sqrt{y-xy}=1$ carefully (using $s=x+y,\,p=xy$) collapses to a perfect square equalling zero, revealing that the relation is nothing but the straight line $x+y=1$. Once we know that, $\dfrac{dy}{dx}=-1$ is immediate, and we just need to find which option reduces to $-1$ on this line.

   Step-by-Step Solution

   1. Square the given equation:

   $x(1-y)+y(1-x)+2\sqrt{xy(1-x)(1-y)} = 1$

   $x+y-2xy+2\sqrt{xy(1-x)(1-y)} = 1$

   2. Let $s=x+y,\ p=xy$. Then:

   $2\sqrt{p(1-x)(1-y)} = 1-s+2p$

   Note $(1-x)(1-y)=1-s+p$. Squaring again:

   $4p(1-s+p) = (1-s+2p)^2$

   Let $q=1-s$. Expanding both sides: LHS $=4pq+4p^2$; RHS $=q^2+4pq+4p^2$. So $0=q^2$, i.e. $q=0$, i.e. $s=1$.

   3. Hence $x+y=1$ is forced — the given relation is the line $y=1-x$ (restricted to the domain where the square roots are real, $0\le x,y\le1$).

   4. Differentiating $y=1-x$ directly: $\dfrac{dy}{dx}=-1$.

   5. Check which option gives $-1$ identically along $y=1-x$:

   • (A): $y-y^2=y(1-y)$. Substituting $y=1-x$: $y(1-y)=(1-x)\cdot x=x-x^2$. So the ratio is exactly $1$, and $-\sqrt1=-1$ for every $x$. ✓ …

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7. 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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8. 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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9. 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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10. 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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11. 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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12. 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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