Q.Find $\frac{dy}{dx}$ in the following: $x = \sin t, y = \cos 2t$

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 $y$.
   4. Differentiate $x^2y^2=-1$ implicitly w.r.t. $x$: $2xy^2+x^2\cdot2y\dfrac{dy}{dx}=0$.
   5. Divide through by $2xy$ (nonzero): $y+x\dfrac{dy}{dx}=0$.
   6. So $\dfrac{dy}{dx}=-\dfrac{y}{x}$.

   Common Mistakes

   • Trying to differentiate both original equations w.r.t. $t$ and eliminate $dt$ directly — doable but far more error-prone than first eliminating $t$ algebraically.
   • Sign slip in the final division step, giving $+y/x$ instead of $-y/x$.

   ✓Final answer

   The correct option is (D) — $-\dfrac{y}{x}$.

   ANSWER: D

2. 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 $\dfrac{dy}{dx}=-\dfrac{1}{x^3 y}$.
   4. At $(-1,2)$: $x^3=(-1)^3=-1$, so $\dfrac{dy}{dx}=-\dfrac{1}{(-1)(2)}=\dfrac{1}{2}$.

   Common Mistakes

   • Getting the sign wrong by mishandling $x^3=-1$ at $x=-1$.
   • Trying to differentiate the arctangents directly without first simplifying, which is far messier.

   ✓Final answer

   The correct option is (C) — $\dfrac{dy}{dx}=\dfrac{1}{2}$.

   ANSWER: C

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, $1+e^{2y}=\dfrac{\pi}{2}\Rightarrow e^{2y}=\dfrac{\pi}{2}-1=\dfrac{\pi-2}{2}$.
   7. Substitute: $\dfrac{dy}{dx} = -\dfrac{3(1)}{2\cdot\frac{\pi-2}{2}} = -\dfrac{3}{\pi-2} = \dfrac{3}{2-\pi}$.

   Common Mistakes

   • Trying to differentiate $\tan$ and $\cot$ directly instead of first converting to a purely algebraic relation between $e^{3x}$ and $e^{2y}$ — this makes implicit differentiation much messier and error-prone.
   • Sign slip when flipping $-\dfrac{3}{\pi-2}$ to $\dfrac{3}{2-\pi}$ (they are equal, but must match the option's form).

   ✓Final answer

   The correct option is (A) — $\dfrac{3}{2-\pi}$.

   ANSWER: A

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. $y = x\tan^2\alpha$.
   5. Since $\tan^2\alpha$ is a constant (does not depend on $x$), $\dfrac{dy}{dx}=\tan^2\alpha = \dfrac{\sin^2\alpha}{\cos^2\alpha}$.

   Common Mistakes

   • Jumping straight into implicit differentiation of the original messy relation instead of spotting the perfect square — much harder and error-prone.
   • Forgetting that $y=x\tan^2\alpha$ makes this literally a line through the origin, so the derivative is just its slope.

   ✓Final answer

   The correct option is (C) — $\dfrac{\sin^2\alpha}{\cos^2\alpha}$.

   ANSWER: C

5. 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:

   $y'=\frac{\frac1{\sqrt2}\left(\sqrt2 a - \frac{2b}{\sqrt2}\right)}{\frac1{\sqrt2}\left(\sqrt2 a + \frac{2b}{\sqrt2}\right)}=\frac{\sqrt2 a-\sqrt2 b}{\sqrt2 a+\sqrt2 b}=\frac{a-b}{a+b}.$

   Common Mistakes

   • Forgetting to expand the product first and instead trying to implicitly differentiate the product form directly, which is far more error-prone.
   • Sign slips when differentiating $\cos x\cos y$ as a product (needs the product rule with $y'$ attached only to the $\cos y$ factor).

   ✓Final answer

   The correct option is (B) — $\dfrac{a - b}{a + b}$.

   ANSWER: B

6. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If $x^2 + y^2 + \sin y = 4$, then the value of $\dfrac{d^2y}{dx^2}$ at $x = -2$ is (A) $-30$ (B) $-34$ (C) $-32$ (D) $-18$

   ›Reveal solutionSolution

   Implicit differentiation of $x^2+y^2+\sin y=4$ twice, using $y(-2)=0$ and $y'(-2)=4$, gives $y''(-2)=-34$.

   Concept and Intuition

   For an implicitly-defined curve, we differentiate the whole equation with respect to $x$ (treating $y$ as a function of $x$ and applying the chain rule to every $y$-term), solve for $y'$, then differentiate the resulting equation again to get $y''$. The key first step is always finding the actual point $(x_0,y_0)$ on the curve, since $y'$ and $y''$ are evaluated there.

   Step-by-Step Solution

   1. Find $y$ at $x=-2$: substituting $x=-2$ into $x^2+y^2+\sin y=4$: $4+y^2+\sin y=4\Rightarrow y^2+\sin y=0$. Clearly $y=0$ satisfies this (and is the relevant branch), so $y(-2)=0$.
   2. First derivative: differentiate $x^2+y^2+\sin y=4$ w.r.t. $x$:

   $2x+2y\,y'+\cos y\cdot y'=0 \implies y'(2y+\cos y)=-2x \implies y'=\frac{-2x}{2y+\cos y}.$

   At $(x,y)=(-2,0)$: $y'=\dfrac{-2(-2)}{2(0)+\cos0}=\dfrac{4}{1}=4$.

   3. Second derivative: differentiate $2x+2y\,y'+\cos y\cdot y'=0$ again w.r.t. $x$, using the product rule on both $2y\,y'$ and $\cos y \cdot y'$:

   $2+2(y')^2+2y\,y''-\sin y\,(y')^2+\cos y\, y''=0.$

   4. Substitute $x=-2$, $y=0$, $y'=4$:

   $2+2(4)^2+2(0)y''-\sin(0)(4)^2+\cos(0)\,y''=0$

   $2+32+0-0+y''=0 \implies y''=-34.$

   Common Mistakes

   • Forgetting to find the actual point $(x_0,y_0)$ first — $y'$ and $y''$ formulas need numeric $y$, not just $x$.
   • Missing a term when differentiating $2y\,y'$ a second time (it needs the product rule: $2(y')^2+2y\,y''$).

   ✓Final answer

   The correct option is (B) — $-34$.

   ANSWER: B

7. 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 $y=\sqrt2,\ u=2$: $\dfrac{du}{dy}=\dfrac{4}{\sqrt2(1-2\log\sqrt2)}=\dfrac{4}{\sqrt2(1-\ln2)}=\dfrac{2\sqrt2}{1-\ln2}$.
   3. Differentiate the nested-log relation $u=\log(x+u)$ w.r.t. $x$: $\dfrac{du}{dx}=\dfrac{1+\frac{du}{dx}}{x+u} \Rightarrow \dfrac{du}{dx}(x+u-1)=1 \Rightarrow \dfrac{du}{dx}=\dfrac1{x+u-1}$. At $x=e^2-2,\ u=2$: $x+u-1=e^2-1$, so $\dfrac{du}{dx}=\dfrac1{e^2-1}$.
   4. Combine via chain rule: since both equal the same $u$, $\dfrac{dy}{dx}=\dfrac{du/dx}{du/dy}=\dfrac{\;\dfrac1{e^2-1}\;}{\dfrac{2\sqrt2}{1-\ln2}}=\dfrac{1-\ln2}{2\sqrt2(e^2-1)}$.

   Common Mistakes

   • Forgetting to verify $u=2$ actually satisfies both self-consistency equations before differentiating.
   • Sign/algebra slips in the implicit differentiation of $u=y^u$ (logarithmic differentiation) — a very common source of error in infinite-tower problems.

   ✓Final answer

   The correct option is (B) — $\dfrac{1 - \log 2}{2\sqrt{2}(e^2 - 1)}$.

   ANSWER: B

8. 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:

   $x\frac{dy}{dx}+y=0 \Rightarrow \frac{dy}{dx}=-\frac{y}{x}$

   5. Then $x^3y\dfrac{dy}{dx}=x^3y\left(-\dfrac{y}{x}\right)=-x^2y^2=-1$ (using step 3).

   Common Mistakes

   • Trying to solve for $x,y$ individually in terms of $t$ (unnecessarily complicated) instead of eliminating $t$ algebraically.
   • Sign error when substituting $dy/dx=-y/x$.

   ✓Final answer

   The correct option is (A) — -1.

   ANSWER: A

9. 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.
   3. Differentiate $xy=\text{constant}$ implicitly: $\dfrac{d}{dx}(xy)=0 \Rightarrow y+x\dfrac{dy}{dx}=0$.
   4. Solve: $\dfrac{dy}{dx}=-\dfrac{y}{x}$.

   Common Mistakes

   • Differentiating the trig terms directly (product/chain rule on $\sin(xy),\cos(xy)$) without noticing the amplitude equals the RHS, missing the much simpler $xy=$const shortcut and getting stuck in messy algebra.
   • Sign error in the implicit derivative of $xy$.

   ✓Final answer

   The correct option is (C) — $\dfrac{-y}{x}$.

   ANSWER: C

10. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.If $x^2+y^2+\sin y=4$, then the value of $\dfrac{d^2y}{dx^2}$ at the point $(-2,0)$ is (A) -34 (B) -32 (C) 34 (D) 32

    ›Reveal solutionSolution

    Implicit differentiation twice on $x^2+y^2+\sin y=4$ gives $y''=-34$ at $(-2,0)$.

    Concept and Intuition

    For an implicitly defined curve, differentiate the whole equation with respect to $x$ once to get $y'$ in terms of $x,y$, then differentiate that resulting equation again (product/chain rule carefully) to isolate $y''$, finally substituting the numeric point.

    Step-by-Step Solution

    1. Differentiate $x^2+y^2+\sin y=4$ w.r.t. $x$:

    $2x+2y\,y'+\cos y\,y'=0 \Rightarrow y'(2y+\cos y)=-2x \Rightarrow y'=\frac{-2x}{2y+\cos y}$

    2. At $(-2,0)$: $y'=\dfrac{-2(-2)}{2(0)+\cos0}=\dfrac{4}{1}=4$.
    3. Differentiate the equation $2x+2y y'+\cos y\, y'=0$ again w.r.t. $x$:

    $2+2(y')^2+2y\,y''-\sin y\,(y')^2+\cos y\,y''=0$

    $2+\left[2-\sin y\right](y')^2+(2y+\cos y)y''=0$

    4. At the point: $y=0$, $y'=4$, $\sin y=0$, $\cos y=1$, so $2y+\cos y=1$:

    $2+(2-0)(16)+1\cdot y''=0 \Rightarrow 2+32+y''=0\Rightarrow y''=-34$

    Common Mistakes

    • Forgetting the $-\sin y\,(y')^2$ term that comes from differentiating $\cos y\,y'$ (product + chain rule).
    • Substituting the point before fully simplifying the second-derivative equation, causing arithmetic slips.

    ✓Final answer

    The correct option is (A) — -34.

    ANSWER: A

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).
    3. At $x=1$: $\dfrac{dy}{dx} = \dfrac{2(1)}{1+1^2} = \dfrac{2}{2} = 1$.

    Common Mistakes

    • Leaving the answer in terms of $e^y$ instead of substituting back $e^y=1+x^2$ to get a numeric slope.
    • Forgetting to evaluate at $x=1$ and stopping at the general formula.

    ✓Final answer

    The correct option is (D) — $1$.

    ANSWER: D

12. 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 $x$: $\cos x = -\sin y\,\dfrac{dy}{dx}$.
    5. Since $\cos y=\sin x$, we have $\sin y=\sqrt{1-\cos^2y}=\sqrt{1-\sin^2x}=\cos x$ (matching branch/sign consistent with the original equation).
    6. So $\cos x = -\cos x\cdot\dfrac{dy}{dx} \Rightarrow \dfrac{dy}{dx}=-1$.

    Common Mistakes

    • Trying to differentiate the square-root terms directly instead of first simplifying the relation — this leads to messy, error-prone algebra.
    • Losing track of sign consistency between $\sin y$ and $\cos x$ when converting one to the other.

    ✓Final answer

    The correct option is (C) — $-1$.

    ANSWER: C