Q.Find $\frac{dy}{dx}$ if $x - y = \pi$.

Showing the 12 most recent of 50 on this concept.

1. 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

2. 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

3. 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

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: $2\cdot\dfrac\pi4 + \dfrac{y'-1}{2} - 2y'\cdot\dfrac\pi4 - \dfrac{1-y'}{2} = 0$.
   4. Combine the two $/2$ terms: $\dfrac{y'-1}{2}-\dfrac{1-y'}{2}=y'-1$.
   5. So: $\dfrac\pi2 + (y'-1) - \dfrac\pi2 y' = 0 \Rightarrow y'\left(1-\dfrac\pi2\right) = 1-\dfrac\pi2 \Rightarrow y'=1$.

   Common Mistakes

   • Sign slips differentiating $\tan^{-1}(x/y)$ with respect to $x$ (the $y'$ appears with a minus sign inside).

   ✓Final answer

   The correct option is (C) — $1$.

   ANSWER: C

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 $x$. ✓
   • (B): at $x=0.3,y=0.7$: $\dfrac{1-y^2}{1-x^2}=\dfrac{0.51}{0.91}\ne1$. ✗
   • (C): at $x=0.3,y=0.7$: $\dfrac{1-y}{1-x}=\dfrac{0.3}{0.7}\ne1$. ✗
   • (D): at $x=0.3,y=0.7$: $\dfrac{x-y}{x+y}=\dfrac{-0.4}{1}<0$, square root undefined. ✗

   Common Mistakes

   • Grinding through raw implicit differentiation without noticing the constraint simplifies to a straight line — leads to messy, error-prone algebra.
   • Checking the options only at the symmetric point $x=y=0.5$, where several options coincidentally also give $-1$ — a second, asymmetric test point is needed to discriminate.
   • Sign error in expanding $(1-s+2p)^2$ vs $4p(1-s+p)$.

   ✓Final answer

   The correct option is (A) — $-\sqrt{\dfrac{y-y^2}{x-x^2}}$.

   ANSWER: A

6. 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: $\dfrac{dy}{dx}(2y-1)=1\Rightarrow\dfrac{dy}{dx}=\dfrac1{2y-1}$.

   Common Mistakes

   • Trying to differentiate the nested radical term-by-term (an infinite process) instead of using the self-similarity trick $y=\sqrt{x+y}$.
   • Sign/rearrangement slip moving $\dfrac{dy}{dx}$ terms to one side.

   ✓Final answer

   The correct option is (D) — $\dfrac{1}{2y-1}$.

   ANSWER: D

7. 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

8. 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. $x$:

   $\frac{s'-1}{s-x}=y's+ys'$

   3. LHS $=(s'-1)(s+x)=\left(\dfrac{x}{s}-1\right)(s+x)=\dfrac{(x-s)(s+x)}{s}=\dfrac{x^2-s^2}{s}=\dfrac{-1}{s}$ (using $s^2-x^2=1$).
   4. RHS $=y's+y\cdot\dfrac{x}{s}$.
   5. So $-\dfrac1s=y's+\dfrac{xy}{s}$. Multiply through by $s$: $-1=y's^2+xy=(1+x^2)y'+xy$.

   Common Mistakes

   • Forgetting $s^2=1+x^2$ when converting $y's^2$ back to $(1+x^2)y'$.
   • Sign slip when simplifying $(s'-1)(s+x)$.

   ✓Final answer

   The correct option is (D) — $-1$.

   ANSWER: D

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. $y'=\dfrac{-(2xy-y^2+3x^2)}{x^2-2xy-3y^2}=\dfrac{y^2-2xy-3x^2}{x^2-2xy-3y^2}$.
   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

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 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

12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.If $y = \log_y x$, then $\dfrac{dy}{dx} =$ ______ (A) $\dfrac{1}{x\log y}$ (B) $\dfrac{\log y}{x(1+\log y)}$ (C) $\dfrac{1}{x(1+\log y)}$ (D) $\dfrac{1}{1+\log y}$

    ›Reveal solutionSolution

    Rewriting $y=\log_y x$ as $y\ln y=\ln x$ and differentiating implicitly gives $\dfrac{dy}{dx}=\dfrac{1}{x(1+\log y)}$.

    Concept and Intuition

    $\log_y x$ means "logarithm of $x$ to base $y$", i.e. $\dfrac{\ln x}{\ln y}$. So the given relation $y=\log_y x$ really means $y=\dfrac{\ln x}{\ln y}$, or equivalently $y\ln y=\ln x$ — a cleaner form to differentiate implicitly, since it avoids a quotient with $y$ in both places.

    Step-by-Step Solution

    1. $y=\log_y x=\dfrac{\ln x}{\ln y} \Rightarrow y\ln y=\ln x$.
    2. Differentiate both sides with respect to $x$, treating $y$ as a function of $x$:

    $\frac{d}{dx}(y\ln y)=\frac{d}{dx}(\ln x)$

    3. LHS (product rule): $\dfrac{dy}{dx}\ln y+y\cdot\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{dy}{dx}(\ln y+1)$.
    4. RHS: $\dfrac1x$.
    5. So $\dfrac{dy}{dx}(\ln y+1)=\dfrac1x \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x(1+\ln y)}=\dfrac{1}{x(1+\log y)}$.

    Common Mistakes

    • Differentiating $y=\dfrac{\ln x}{\ln y}$ directly as a quotient (messier and error-prone) instead of first cross-multiplying to $y\ln y=\ln x$.
    • Forgetting the extra $\dfrac{dy}{dx}$ term that arises from differentiating $\ln y$ (chain rule) when applying the product rule to $y\ln y$.

    ✓Final answer

    The correct option is (C) — $\dfrac{1}{x(1+\log y)}$.

    ANSWER: C