Q.Find $\frac{dy}{dx}$ in the following: $(5x)^{3 \cos 2x}$

Showing the 12 most recent of 50 on this concept.

1. 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

2. 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

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 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

5. 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

6. 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

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 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

9. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.If $x^3 - 2x^2y^2 + 5x + y - 5 = 0$, then at $(1,1)$, $y''(1) =$ (A) $-\frac{197}{27}$ (B) $\frac{125}{31}$ (C) $12$ (D) $-\frac{238}{27}$

   ›Reveal solutionSolution

   Implicit differentiation twice, using the values at $(1,1)$ and the first-derivative value $y'(1)=4/3$, gives $y''(1) = -238/27$.

   Concept and Intuition

   For a curve defined implicitly by $F(x,y)=0$, we differentiate throughout with respect to $x$, treating $y$ as a function of $x$ (chain rule at every $y$-term), to get an equation involving $y'$; solving that gives $y'$ in terms of $x,y$. Differentiating that resulting equation once more (again using the chain rule, now also needing $y'$ itself) and substituting known values gives $y''$.

   Step-by-Step Solution

   1. Verify $(1,1)$ lies on the curve: $1-2+5+1-5=0$. ✓
   2. Differentiate $x^3-2x^2y^2+5x+y-5=0$ w.r.t. $x$: $3x^2 - 2(2xy^2+2x^2yy') + 5+y' = 0 \Rightarrow 3x^2-4xy^2-4x^2yy'+5+y'=0$.
   3. Collect $y'$ terms: $y'(1-4x^2y) = -3x^2+4xy^2-5$.
   4. At $(1,1)$: numerator $=-3(1)+4(1)(1)-5=-4$; denominator $=1-4(1)(1)=-3$. So $y'(1)=\dfrac{-4}{-3}=\dfrac43$.
   5. Differentiate $y'(1-4x^2y)=-3x^2+4xy^2-5$ again w.r.t. $x$ (product rule on the LHS, chain rule throughout): LHS derivative: $y''(1-4x^2y) + y'\cdot\big(-(8xy+4x^2y')\big)$. RHS derivative: $-6x+4(y^2+2xyy')$.
   6. At $(1,1)$ with $y'=4/3$: LHS becomes $y''(-3) + \tfrac43\left(-(8+\tfrac{16}{3})\right) = -3y'' - \tfrac43\cdot\tfrac{40}{3} = -3y'' - \tfrac{160}{9}$. RHS becomes $-6+4\left(1+\tfrac83\right) = -6+4\cdot\tfrac{11}{3} = -6+\tfrac{44}{3}=\tfrac{26}{3}$.
   7. So $-3y'' - \tfrac{160}9 = \tfrac{26}3 = \tfrac{78}9 \Rightarrow -3y'' = \tfrac{78+160}{9}=\tfrac{238}9 \Rightarrow y'' = -\dfrac{238}{27}$.

   Common Mistakes

   • Losing track of the product-rule terms when differentiating the already-differentiated (first-derivative) equation a second time.
   • Arithmetic slips converting fractions with denominators $3$ and $9$ together.

   ✓Final answer

   The correct option is (D) — $-\dfrac{238}{27}$.

   ANSWER: D

10. 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

11. 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

12. 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