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

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.
   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

2. 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

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 $x$: $\cos x = -\sin y\,\dfrac{dy}{dx}$.
   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

4. 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

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 $\dfrac1y \dfrac{dy}{dx} = \cos x\,\log(\tan x) + \sec x$.
   5. Multiply by $y = (\tan x)^{\sin x}$: $\dfrac{dy}{dx} = (\tan x)^{\sin x}\left\{\sec x + \cos x\,\log(\tan x)\right\}$.

   Common Mistakes

   • Sign error on the second term (writing $-\cos x\log(\tan x)$ instead of $+$), which would incorrectly point to option (C).
   • Forgetting to multiply back by $y$ at the end and leaving the answer as just $y'/y$.

   ✓Final answer

   The correct option is (A) — $(\tan x)^{\sin x}\{\sec x+(\cos x)(\log(\tan x))\}$.

   ANSWER: A

6. 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

   $y' = y\left[\frac{x\cot x}{\log\sin x} + \log(\log\sin x) - \frac{1}{\log x} - \log(\log x)\right],$

   which is exactly option (C).

   Common Mistakes

   • Sign errors when differentiating $\ln(\ln x)$ vs $1/\ln x$ terms — easy to drop or flip a minus sign.
   • Forgetting the quotient rule inside $u'$ (treating $\ln\sin x$ and $\ln x$ as independent rather than a ratio).

   ✓Final answer

   The correct option is (C) — $y\left[\dfrac{x\cot x}{\log\sin x}+\log(\log\sin x)-\dfrac{1}{\log x}-\log(\log x)\right]$.

   ANSWER: C

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:

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

   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

8. 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

9. 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

10. 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

11. 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. $y'=\dfrac{-(2xy-y^2+3x^2)}{x^2-2xy-3y^2}=\dfrac{y^2-2xy-3x^2}{x^2-2xy-3y^2}$.
    4. At $(1,1)$: numerator $=1-2-3=-4$; denominator $=1-2-3=-4$.
    5. $y'=\dfrac{-4}{-4}=1$.

    Common Mistakes

    • Sign slips when moving terms across the equation before isolating $y'$.
    • Forgetting the product rule on the mixed term $-xy^2$ (it needs both a $y^2$ contribution and a $2xyy'$ contribution).

    ✓Final answer

    The correct option is (A) — $1$.

    ANSWER: A

12. 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