Q.Find $\frac{dy}{dx}$ in the following: $x^3 + x^2y + xy^2 + y^3 = 81$

Showing the 12 most recent of 50 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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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 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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9. 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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10. 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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11. 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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12. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.The length of the normal drawn to the curve $2x^3 + 2y^3 = 9xy$ at the point $(2, 1)$, is (A) $\dfrac{\sqrt{41}}{4}$ (B) $\dfrac{2}{3}\sqrt{41}$ (C) $\sqrt{5}$ (D) $\dfrac{2}{3}\sqrt{5}$

    ›Reveal solutionSolution

    Find the tangent slope by implicit differentiation, then apply the standard formula for the length of the normal at a point on a curve.

    Concept and Intuition

    The length of the normal segment (from the curve point down to where the normal line meets the $x$-axis) is $|y_1|\sqrt{1+m^2}$, where $m$ is the slope of the tangent at that point — this comes from the right triangle formed by the ordinate, the subnormal, and the normal itself.

    Step-by-Step Solution

    1. Differentiate $2x^3+2y^3=9xy$ implicitly: $6x^2+6y^2y'=9(y+xy')$, i.e. $6x^2+6y^2y'=9y+9xy'$.
    2. Collect: $y'(6y^2-9x)=9y-6x^2\Rightarrow y'=\dfrac{9y-6x^2}{6y^2-9x}$.
    3. At $(2,1)$: numerator $=9(1)-6(4)=9-24=-15$; denominator $=6(1)-9(2)=6-18=-12$.
    4. $y'=\dfrac{-15}{-12}=\dfrac{5}{4}=m$. …

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