Q.If $y = (\tan^{-1} x)^2$, show that $(x^2+1)^2 y_2 + 2x(x^2+1) y_1 = 2$. Miscellaneous Examples

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If $y = \sin^{-1}x$ then $(1-x^2)y_2 - xy_1 =$ (A) $0$ (B) $1$ (C) $2$ (D) $2y$

   ›Reveal solutionSolution

   $y=\sin^{-1}x$ satisfies $(1-x^2)y_2 - xy_1 = 0$ — a classic second-order ODE identity.

   Concept and Intuition

   Differentiating $y=\sin^{-1}x$ once gives $y_1=\dfrac{1}{\sqrt{1-x^2}}$, i.e. $(1-x^2)y_1^2=1$. Differentiating this relation again (implicitly) produces the required second-order identity — a standard result worth memorizing.

   Step-by-Step Solution

   1. $y_1 = \dfrac{1}{\sqrt{1-x^2}} \Rightarrow (1-x^2)y_1^2 = 1$.
   2. Differentiate both sides w.r.t. $x$: $-2x y_1^2 + (1-x^2)\cdot 2y_1y_2 = 0$.
   3. Divide by $2y_1$ (nonzero): $-xy_1 + (1-x^2)y_2 = 0$. …

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2. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If $y=(\tan^{-1}2x)^2+(\cot^{-1}2x)^2$, then $(1+4x^2)^2y''-16=$ (A) $8xy'$ (B) $-8x(1+4x^2)y'$ (C) $8x(1+4x^2)y'$ (D) $-8xy'$

   ›Reveal solutionSolution

   Using $\tan^{-1}2x+\cot^{-1}2x=\pi/2$ collapses $y$ to a quadratic in $a=\tan^{-1}2x$; two differentiations and elimination of $a$ give the required identity, matching option (B).

   Concept and Intuition

   Whenever a problem gives $(\tan^{-1}u)^2+(\cot^{-1}u)^2$, exploit the constant-sum identity $\tan^{-1}u+\cot^{-1}u=\pi/2$ to reduce two inverse-trig terms to one variable. This turns a messy-looking function into a clean quadratic, after which ordinary calculus (product/chain rule) finishes the job.

   Step-by-Step Solution

   1. Let $a=\tan^{-1}(2x)$. Then $\cot^{-1}(2x)=\dfrac\pi2-a$, so $y=a^2+\left(\dfrac\pi2-a\right)^2=2a^2-\pi a+\dfrac{\pi^2}{4}$.
   2. $\dfrac{da}{dx}=\dfrac{2}{1+4x^2}$. So $y'=(4a-\pi)\cdot\dfrac{2}{1+4x^2}=\dfrac{2(4a-\pi)}{D}$, where $D=1+4x^2$.
   3. Let $N=4a-\pi$, so $y'=2N/D$. Differentiate again: $\dfrac{dN}{dx}=4\cdot\dfrac{2}{D}=\dfrac{8}{D}$, and $\dfrac{dD}{dx}=8x$.
   4. $y''=2\cdot\dfrac{N'D-ND'}{D^2}=2\cdot\dfrac{8-8xN}{D^2}=\dfrac{16-16xN}{D^2}$, so $D^2y''=16-16xN$. …

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3. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $y = (ax + b)\cos x$, then $y_2 + y_1\sin 2x + y(1 + \sin^2 x) =$ (A) $y_2 \cos^2 x$ (B) $y_2 \sin^2 x$ (C) $y_1 \sin^2 x$ (D) $y\sin^2 x$

   ›Reveal solutionSolution

   Direct computation of $y_1,y_2$ from $y=(ax+b)\cos x$ and substitution shows the given combination collapses exactly to $y_2\sin^2x$.

   Concept and Intuition

   This is a standard "verify the differential relation" problem: compute the first and second derivatives explicitly, then substitute back into the given combination and simplify using $\sin^2x+\cos^2x=1$ until only a multiple of $y_2$ (or another named quantity) survives.

   Step-by-Step Solution

   1. $y=(ax+b)\cos x$.
   2. $y_1=a\cos x-(ax+b)\sin x$ (product rule).
   3. $y_2=-a\sin x-\big[a\sin x+(ax+b)\cos x\big]=-2a\sin x-(ax+b)\cos x=-2a\sin x-y$.
   4. Since $y=(ax+b)\cos x$, we have $ax+b=\dfrac{y}{\cos x}$ (for $\cos x\ne0$).
   5. $y_1\sin2x=[a\cos x-(ax+b)\sin x]\cdot2\sin x\cos x=2a\sin x\cos^2x-2(ax+b)\sin^2x\cos x=2a\sin x\cos^2x-2y\sin^2x$.
   6. Now sum: $y_2+y_1\sin2x+y(1+\sin^2x)$ $=(-2a\sin x-y)+(2a\sin x\cos^2x-2y\sin^2x)+y+y\sin^2x$ $=-2a\sin x(1-\cos^2x)+(-y+y)+(-2y\sin^2x+y\sin^2x)$ $=-2a\sin^3x-y\sin^2x$. …

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4. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If $y=\tan(\log x)$, then $\dfrac{d^2y}{dx^2}=$ (A) $\dfrac{-\sec^2(\log x)[1+2\tan x]}{x^2}$ (B) $\dfrac{\sec^2(\log x)[1+\tan(\log x)]}{x^2}$ (C) $\dfrac{\sec(\log x)[2\tan(\log x)-1]}{x^2}$ (D) $\dfrac{\sec^2(\log x)[2\tan(\log x)-1]}{x^2}$

   ›Reveal solutionSolution

   Repeated chain-rule differentiation of $\tan(\log x)$; the trick is differentiating $\sec^2(\log x)/x$ correctly with the quotient rule.

   Concept and Intuition

   Every time we differentiate a function of $\log x$, a factor of $1/x$ appears from the chain rule. So the second derivative naturally produces a $1/x^2$ term, and we must track how the inner derivative of $\sec^2(\log x)$ itself brings down another factor.

   Step-by-Step Solution

   1. $y=\tan(\log x)\Rightarrow y'=\sec^2(\log x)\cdot\dfrac1x$.
   2. Write $y'=\dfrac{\sec^2(\log x)}{x}$ and differentiate using the quotient rule:

   $y''=\frac{x\cdot\frac{d}{dx}\sec^2(\log x)-\sec^2(\log x)}{x^2}$

   3. $\dfrac{d}{dx}\sec^2(\log x)=2\sec(\log x)\cdot\sec(\log x)\tan(\log x)\cdot\dfrac1x=\dfrac{2\sec^2(\log x)\tan(\log x)}{x}$.
   4. So $x\cdot\dfrac{d}{dx}\sec^2(\log x)=2\sec^2(\log x)\tan(\log x)$. …

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5. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.If $y = \tan(3\tan^{-1} x)$, then $(1 - 3x^2)\dfrac{d^2y}{dx^2} - 12x\dfrac{dy}{dx} =$ (A) $6(x + y)$ (B) $6(y - x)$ (C) $6y$ (D) $-6x$

   ›Reveal solutionSolution

   Writing $y=\tan(3\tan^{-1}x)$ explicitly as the rational function $\dfrac{3x-x^3}{1-3x^2}$ and computing $y',y''$ directly shows that $(1-3x^2)y''-12xy'$ simplifies exactly to $6(y-x)$.

   Concept and Intuition

   Using the triple-angle identity for tangent, $\tan(3\theta)=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$, with $\theta=\tan^{-1}x$ (so $\tan\theta=x$), the whole expression becomes an explicit rational function of $x$. Rather than working abstractly with the differential equation, it is often faster to differentiate this closed form directly and simplify the target combination.

   Step-by-Step Solution

   1. $y=\tan(3\tan^{-1}x)=\dfrac{3x-x^3}{1-3x^2}$ (triple-angle formula for tangent).
   2. Differentiate using the quotient rule: with $N=3x-x^3,\,N'=3-3x^2$ and $D=1-3x^2,\,D'=-6x$, $y'=\dfrac{N'D-ND'}{D^2}=\dfrac{(3-3x^2)(1-3x^2)+6x(3x-x^3)}{(1-3x^2)^2}$. Expanding the numerator gives $3(1+x^2)^2$, so $y'=\dfrac{3(1+x^2)^2}{(1-3x^2)^2}$.
   3. Differentiate again similarly to get $y''=\dfrac{48x(1+x^2)}{(1-3x^2)^3}$.
   4. Compute $(1-3x^2)y''=\dfrac{48x(1+x^2)}{(1-3x^2)^2}$ and $12xy'=\dfrac{36x(1+x^2)^2}{(1-3x^2)^2}$.
   5. Subtracting: $(1-3x^2)y''-12xy'=\dfrac{12x(1+x^2)\left[4-3(1+x^2)\right]}{(1-3x^2)^2}=\dfrac{12x(1+x^2)(1-3x^2)}{(1-3x^2)^2}=\dfrac{12x(1+x^2)}{1-3x^2}$. …

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6. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.If $\dfrac{d^n y}{dx^n} = y_n$ and $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$, then $4xy_2 + 2y_1 =$ (A) $-y$ (B) $y$ (C) $2y$ (D) $-2y$

   ›Reveal solutionSolution

   Differentiating $y=e^{\sqrt x}+e^{-\sqrt x}$ twice (using the chain rule through $u=\sqrt x$) and simplifying shows that the combination $4xy_2+2y_1$ collapses exactly back to $y$ — this is the differential equation this function satisfies.

   Concept and Intuition

   Functions built from $e^{\pm\sqrt x}$ satisfy a specific second-order ODE because of how the chain rule through $u=\sqrt{x}$ introduces extra factors of $x$ and $\sqrt x$ each time we differentiate. Recognizing the pattern $4xy_2+2y_1$ as "the ODE this family solves" is a common Class-12 differentiation-application question.

   Step-by-Step Solution

   1. Let $u=\sqrt x$, so $y=e^{u}+e^{-u}$ and $\dfrac{du}{dx}=\dfrac{1}{2\sqrt x}$.
   2. First derivative: $y_1=\dfrac{dy}{du}\cdot\dfrac{du}{dx}=(e^{u}-e^{-u})\cdot\dfrac{1}{2\sqrt x}=\dfrac{e^{\sqrt x}-e^{-\sqrt x}}{2\sqrt x}$.
   3. Write $y_1=\dfrac{v}{w}$ with $v=e^{\sqrt x}-e^{-\sqrt x}$ and $w=2\sqrt x$. Then $\dfrac{dv}{dx}=\dfrac{e^{\sqrt x}+e^{-\sqrt x}}{2\sqrt x}=\dfrac{y}{2\sqrt x}$ and $\dfrac{dw}{dx}=\dfrac{1}{\sqrt x}$.
   4. Quotient rule: $y_2=\dfrac{w\cdot v' - v\cdot w'}{w^2}=\dfrac{2\sqrt x\cdot\frac{y}{2\sqrt x} - v\cdot\frac{1}{\sqrt x}}{4x}=\dfrac{y-\frac{v}{\sqrt x}}{4x}$. …

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7. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots \infty}}}$ then the value of $\dfrac{d^2y}{dx^2}$ at the point $(\pi, 1)$ is (A) $2$ (B) $-2$ (C) $-\dfrac{1}{2}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Convert the infinite nested radical into an implicit algebraic equation $y^2-y=\sin x$, then differentiate twice implicitly and substitute the given point.

   Concept and Intuition

   An infinite nested radical $y=\sqrt{\sin x+\sqrt{\sin x+\cdots}}$ satisfies the self-referential relation $y=\sqrt{\sin x+y}$ (the expression under the first root is $\sin x$ plus the same infinite tail $y$). Squaring converts this into a clean implicit equation that can be differentiated normally.

   Step-by-Step Solution

   1. $y=\sqrt{\sin x+y} \Rightarrow y^2=\sin x+y \Rightarrow y^2-y=\sin x$.
   2. Differentiate w.r.t. $x$: $2yy'-y'=\cos x \Rightarrow y'(2y-1)=\cos x$.
   3. At $(\pi,1)$: $\cos\pi=-1$, $2(1)-1=1$, so $y'(1)=-1 \Rightarrow y'=-1$. …

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8. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If $y = x\log\left(\dfrac{1}{ax}+\dfrac{1}{a}\right)$, then $x(x+1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}-y=$ (A) $0$ (B) $1+x$ (C) $-1$ (D) $x$

   ›Reveal solutionSolution

   Writing $y=x\log(1+x)-x\log a-x\log x$ and computing $y',y''$ directly, the combination $x(x+1)y''+xy'-y$ collapses to the constant $-1$.

   Concept and Intuition

   Simplify the logarithm first (combine the fractions inside), then this becomes a standard "compute derivatives and substitute into a differential expression" problem — careful term-by-term bookkeeping is the key, not a clever trick.

   Step-by-Step Solution

   1. $\dfrac{1}{ax}+\dfrac1a=\dfrac{1+x}{ax}$, so $y=x\log(1+x)-x\log a-x\log x$.
   2. $y'=\log(1+x)+\dfrac{x}{1+x}-\log a-\log x-1$.
   3. $y''=\dfrac{1}{1+x}+\dfrac{1}{(1+x)^2}-\dfrac1x$.
   4. $x(x+1)y''=x(x+1)\left[\dfrac1{1+x}+\dfrac1{(1+x)^2}-\dfrac1x\right]=x+\dfrac{x}{1+x}-(x+1)=\dfrac{x}{1+x}-1$.
   5. $xy'=x\log(1+x)+\dfrac{x^2}{1+x}-x\log a-x\log x-x$.
   6. $xy'-y=\left[x\log(1+x)+\dfrac{x^2}{1+x}-x\log a-x\log x-x\right]-\left[x\log(1+x)-x\log a-x\log x\right]=\dfrac{x^2}{1+x}-x$. …

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9. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $\sqrt{x} + \sqrt{y} = \sqrt{a}$, then $\left(\dfrac{d^2y}{dx^2}\right)_{x = a} =$ (A) $\dfrac{1}{a}$ (B) $\dfrac{1}{2a}$ (C) $\dfrac{1}{2\sqrt{a}}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Solve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ explicitly for $y$ as a function of $x$ and differentiate twice; the answer is $\dfrac{1}{2a}$.

   Concept and Intuition

   Instead of doing messy implicit differentiation (which produces a $y^{-1/2}$ term that blows up as $y\to0$), it is far cleaner here to isolate $y$ first, since the given relation is easy to invert. This turns an implicit-differentiation problem into an ordinary one.

   Step-by-Step Solution

   1. From $\sqrt{x}+\sqrt{y}=\sqrt{a}$, we get $\sqrt{y}=\sqrt{a}-\sqrt{x}$, so

   $y=(\sqrt a-\sqrt x)^2=a+x-2\sqrt{ax}=a+x-2\sqrt a\,x^{1/2}.$

   2. Differentiate once:

   $\frac{dy}{dx}=1-2\sqrt a\cdot\frac12 x^{-1/2}=1-\sqrt a\,x^{-1/2}=1-\sqrt{\frac{a}{x}}.$

   3. Differentiate again:

   $\frac{d^2y}{dx^2}=-\sqrt a\left(-\frac12\right)x^{-3/2}=\frac{\sqrt a}{2}x^{-3/2}.$

   4. Substitute $x=a$: …

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10. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If $y = \dfrac{\alpha x + \beta}{\gamma x + \delta}$, then $2y_1 y_3 =$ (A) $2y_2^{\,3}$ (B) $3y_2^{\,2}$ (C) $y_2^{\,2}$ (D) $3y_3^{\,2}$

    ›Reveal solutionSolution

    For a Möbius (bilinear) function, each successive derivative is proportional to the previous, letting us relate $y_1,y_2,y_3$ purely algebraically.

    Concept and Intuition

    A function $y=\dfrac{\alpha x+\beta}{\gamma x+\delta}$ always simplifies so its first derivative is $k(\gamma x+\delta)^{-2}$ for a constant $k$ depending on $\alpha,\beta,\gamma,\delta$. Every higher derivative is then just this same power function times a numeric constant, which makes ratios like $y_1y_3/y_2^2$ pure numbers.

    Step-by-Step Solution

    1. Write $y=\dfrac{\alpha x+\beta}{\gamma x+\delta}=\dfrac{\alpha}{\gamma}+\dfrac{k}{\gamma(\gamma x+\delta)}$ where $k=\alpha\delta-\beta\gamma$ (standard identity: $y'=\dfrac{\alpha\delta-\beta\gamma}{(\gamma x+\delta)^2}$).
    2. $y_1=k(\gamma x+\delta)^{-2}$.
    3. $y_2=\dfrac{d}{dx}y_1=-2k\gamma(\gamma x+\delta)^{-3}$.
    4. $y_3=\dfrac{d}{dx}y_2=6k\gamma^2(\gamma x+\delta)^{-4}$. …

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11. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.If $x^3+y^3=3axy$, then at $\left(\dfrac{3a}{2},\dfrac{3a}{2}\right)$ the value of $3ay''+40$ is (A) $-5$ (B) $0$ (C) $8$ (D) $1$

    ›Reveal solutionSolution

    Two rounds of implicit differentiation on the folium-type curve give $y'=-1$ and $y''=-\dfrac{32}{3a}$ at the symmetric point, so $3ay''+40=8$.

    Concept and Intuition

    For implicit curves like $x^3+y^3=3axy$, differentiate once to get $y'$ in terms of $x,y$, evaluate at the point, then differentiate the resulting relation (which still contains $y'$) once more to isolate $y''$ — substituting the already-known $y'$ value at that stage keeps the algebra manageable.

    Step-by-Step Solution

    1. Differentiate $x^3+y^3=3axy$: $3x^2+3y^2y'=3a(y+xy')$, i.e. $x^2+y^2y'=a(y+xy')$ (divided by 3).
    2. Solve for $y'$: $y^2y'-axy'=ay-x^2 \Rightarrow y'=\dfrac{ay-x^2}{y^2-ax}$.
    3. At $\left(\dfrac{3a}2,\dfrac{3a}2\right)$: numerator $=a\cdot\dfrac{3a}2-\left(\dfrac{3a}2\right)^2=\dfrac{3a^2}2-\dfrac{9a^2}4=-\dfrac{3a^2}4$; denominator $=\left(\dfrac{3a}2\right)^2-a\cdot\dfrac{3a}2=\dfrac{9a^2}4-\dfrac{3a^2}2=\dfrac{3a^2}4$. So $y'=-1$.
    4. From $x^2+y^2y'=ay+axy'$ (i.e. $y^2y'-axy'=ay-x^2$), differentiate again: $2y(y')^2+y^2y''-ay'-axy''=ay'-2x$.
    5. Rearranged: $y''(y^2-ax)=2ay'-2x-2y(y')^2$. …

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12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.If $y=1+x+x^2+x^3+\ldots\infty$ and $|x|<1$, then $y''=$ (A) $2yy'$ (B) $\dfrac{2y}{y'}$ (C) $\dfrac{y'}{2y}$ (D) $2y^2y'$

    ›Reveal solutionSolution

    Sum the geometric series to $y=\dfrac{1}{1-x}$, differentiate twice, and express the result in terms of $y,y'$: it is $2yy'$.

    Concept and Intuition

    The infinite series $1+x+x^2+\cdots$ is geometric with ratio $x$, converging (for $|x|<1$) to $\dfrac{1}{1-x}$. Once $y$ has this closed form, its derivatives are straightforward power-rule computations, and both $y'$ and $y''$ turn out to be powers of $(1-x)^{-1}$, so they can be re-expressed purely in terms of $y$ and $y'$.

    Step-by-Step Solution

    1. $y=1+x+x^2+x^3+\cdots=\dfrac{1}{1-x}=(1-x)^{-1}$.
    2. $y'=(1-x)^{-2}$. Notice $y^2=(1-x)^{-2}=y'$, i.e. $y'=y^2$.
    3. $y''=2(1-x)^{-3}$.
    4. Write $(1-x)^{-3}=(1-x)^{-1}\cdot(1-x)^{-2}=y\cdot y'$. …

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