Q.Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: $x + y = \tan^{-1} y$ : $y^2 y' + y^2 + 1 = 0$

Showing the 12 most recent of 50 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.If $x^2+y^2=1$, then ______. (A) $y(y'')-4(y')^2+1=0$ (B) $y(y'')+(y')^2+1=0$ (C) $y(y'')-(y')^2-1=0$ (D) $y(y'')+2(y')^2+1=0$

   ›Reveal solutionSolution

   Differentiating the circle equation $x^2+y^2=1$ twice implicitly gives the differential equation $yy''+(y')^2+1=0$.

   Concept and Intuition

   Any implicit curve, when differentiated repeatedly with respect to $x$ treating $y$ as a function of $x$, yields a differential equation that the curve satisfies. Here we just need to differentiate twice and simplify.

   Step-by-Step Solution

   1. Start with $x^2+y^2=1$.
   2. Differentiate with respect to $x$: $2x+2yy'=0 \implies x+yy'=0$.
   3. Differentiate again with respect to $x$: $1 + (y')^2 + y y'' = 0$ (using the product rule on $yy'$, which gives $y'\cdot y' + y\cdot y''$).
   4. So the relation is $y y'' + (y')^2 + 1 = 0$. …

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2. 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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3. 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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4. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $x^2 \tan^{-1}\dfrac{y}{x} - y^2 \tan^{-1}\dfrac{x}{y} = k$, then $\left(\dfrac{dy}{dx}\right)_{(1,1)} =$ (A) $0$ (B) $\pi/4$ (C) $1$ (D) $\pi/2$

   ›Reveal solutionSolution

   Implicit differentiation of $x^2\tan^{-1}(y/x)-y^2\tan^{-1}(x/y)=k$ evaluated at $(1,1)$ gives $dy/dx=1$.

   Concept and Intuition

   Differentiate both terms using product and chain rules, being careful with the derivatives of $\tan^{-1}(y/x)$ and $\tan^{-1}(x/y)$ with respect to $x$ (treating $y$ as a function of $x$).

   Step-by-Step Solution

   1. $\dfrac{d}{dx}\left[x^2\tan^{-1}\dfrac{y}{x}\right] = 2x\tan^{-1}\dfrac{y}{x} + x^2\cdot\dfrac{y'x-y}{x^2+y^2}$.
   2. $\dfrac{d}{dx}\left[y^2\tan^{-1}\dfrac{x}{y}\right] = 2yy'\tan^{-1}\dfrac{x}{y} + y^2\cdot\dfrac{y-xy'}{x^2+y^2}$.
   3. Setting derivative of LHS $=0$ and evaluating at $(1,1)$, where $\tan^{-1}(1)=\pi/4$ for both terms: $2\cdot\dfrac\pi4 + \dfrac{y'-1}{2} - 2y'\cdot\dfrac\pi4 - \dfrac{1-y'}{2} = 0$. …

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5. 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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6. 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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7. 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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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 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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10. 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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11. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.If the tangent drawn to the curve $(x^2+1)(y-3)=x$ at a point P, lying in the first quadrant, is a horizontal line, then the equation of the normal at the point P is (A) $x=\dfrac{7}{2}$ (B) $x=1$ (C) $y=\dfrac{7}{2}$ (D) $y=1$

    ›Reveal solutionSolution

    This tests implicit differentiation plus the geometric fact that a horizontal tangent forces a vertical normal. The answer is $x=1$.

    Concept and Intuition

    The tangent and normal at a point are always perpendicular. If the tangent line is horizontal (slope $0$), the normal must be vertical (undefined slope) — a line of the form $x=\text{constant}$ passing through the point of tangency. So really the whole problem reduces to finding the point $P$ where $\dfrac{dy}{dx}=0$.

    Step-by-Step Solution

    1. Solve the given curve for $y$: from $(x^2+1)(y-3)=x$, we get $y=3+\dfrac{x}{x^2+1}$.
    2. Differentiate: $\dfrac{dy}{dx}=\dfrac{(x^2+1)(1)-x(2x)}{(x^2+1)^2}=\dfrac{1-x^2}{(x^2+1)^2}$.
    3. Set the tangent slope to zero (horizontal tangent): $1-x^2=0\Rightarrow x=\pm1$.
    4. Since $P$ lies in the first quadrant, take $x=1$. Then $y=3+\dfrac{1}{1+1}=3+\dfrac12=\dfrac72$. So $P=\left(1,\dfrac72\right)$, which indeed has both coordinates positive. …

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12. 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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