Q.Differentiate $\sqrt{\dfrac{(x-3)(x^2+4)}{3x^2+4x+5}}$ w.r.t. $x$.

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If $x = \cos^3 \theta - \sin^3 \theta$ and $y = \sqrt[3]{\cos \theta} - \sqrt[3]{\sin \theta}$, then the value of $\dfrac{dy}{dx}$ at $\theta = \dfrac{\pi}{4}$ is (A) $\dfrac{2}{9} 3\sqrt{2}$ (B) $\dfrac{3\sqrt{2}}{3}$ (C) $\dfrac{4}{9} 3\sqrt{2}$ (D) $\dfrac{3\sqrt{2}}{9}$

   ›Reveal solutionSolution

   Differentiate $x$ and $y$ separately with respect to the parameter $\theta$ and divide. At $\theta=\pi/4$ this gives $\dfrac{dy}{dx}=\dfrac{4}{9}\cdot2^{-2/3}=\dfrac{2}{9}\sqrt[3]{2}$ — option (A).

   The concept first

   When both coordinates are given through a parameter, $x=x(\theta)$ and $y=y(\theta)$, the chain rule gives

   $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}\qquad\left(\text{provided } \frac{dx}{d\theta}\neq0\right).$

   Never try to eliminate $\theta$ here — with a cube and a cube root in the same problem that would be brutal. Just differentiate each expression in $\theta$ and take the quotient at the required value.

   The symmetry $\sin\frac\pi4=\cos\frac\pi4=\frac{1}{\sqrt2}$ makes the arithmetic collapse very neatly, so keep the powers of $2$ in index form until the end.

   Step-by-step

   1. Differentiate $x=\cos^3\theta-\sin^3\theta$:

   $\frac{dx}{d\theta}=3\cos^2\theta(-\sin\theta)-3\sin^2\theta(\cos\theta)=-3\sin\theta\cos\theta\left(\cos\theta+\sin\theta\right).$

   2. Differentiate $y=\cos^{1/3}\theta-\sin^{1/3}\theta$:

   $\frac{dy}{d\theta}=\frac13\cos^{-2/3}\theta\,(-\sin\theta)-\frac13\sin^{-2/3}\theta(\cos\theta) =-\frac13\left(\sin\theta\,\cos^{-2/3}\theta+\cos\theta\,\sin^{-2/3}\theta\right).$

   3. Put $\theta=\dfrac{\pi}{4}$, where $\sin\theta=\cos\theta=2^{-1/2}$:

   $\frac{dx}{d\theta}=-3\left(\frac12\right)\left(\frac{2}{\sqrt2}\right)=-3\cdot\frac12\cdot\sqrt2=-\frac{3\sqrt2}{2}.$

   $\frac{dy}{d\theta}=-\frac13\left(2\cdot 2^{-1/2}\cdot 2^{1/3}\right)=-\frac23\,2^{-1/2+1/3}=-\frac23\,2^{-1/6}.$

   (Here $\cos^{-2/3}\theta=\left(2^{-1/2}\right)^{-2/3}=2^{1/3}$, and the two terms are identical, hence the factor $2$.)

   4. Divide:

   $\frac{dy}{dx}=\frac{-\frac23\,2^{-1/6}}{-\frac{3}{2}\,2^{1/2}}=\frac{2}{3}\cdot\frac{2}{3}\cdot 2^{-1/6-1/2}=\frac49\,2^{-2/3}.$

   5. Tidy the surd. Since $2^{-2/3}=\dfrac{2^{1/3}}{2}$,

   $\frac{dy}{dx}=\frac49\cdot\frac{2^{1/3}}{2}=\frac{2}{9}\,2^{1/3}=\frac{2}{9}\sqrt[3]{2}\approx 0.28 .$

   ✓Final answer

   $\left.\dfrac{dy}{dx}\right|_{\theta=\pi/4}=\dfrac{2}{9}\sqrt[3]{2}$.

   ANSWER: A

2. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $x = \sin 2\theta \cos 3\theta$, $y = \sin 3\theta \cos 2\theta$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{2\cos 5\theta + \cos 3\theta \cos 2\theta}{2\cos 5\theta - \sin 3\theta \sin 2\theta}$ (B) $\dfrac{2\cos 5\theta + \sin 3\theta \sin 2\theta}{2\cos 5\theta + \cos 3\theta \cos 2\theta}$ (C) $\dfrac{2\cos 5\theta - \sin 3\theta \sin 2\theta}{2\cos 5\theta + \cos 3\theta \cos 2\theta}$ (D) $\dfrac{2\cos 5\theta - \cos 3\theta \cos 2\theta}{2\cos 5\theta - \sin 3\theta \sin 2\theta}$

   ›Reveal solutionSolution

   Parametric differentiation gives $\dfrac{dy}{dx}=\dfrac{2\cos5\theta+\cos3\theta\cos2\theta}{2\cos5\theta-\sin3\theta\sin2\theta}$. Option (A).

   Solution

   With $x=\sin2\theta\cos3\theta$ and $y=\sin3\theta\cos2\theta$, differentiate each by the product rule:

   $\frac{dx}{d\theta}=2\cos2\theta\cos3\theta-3\sin2\theta\sin3\theta,\qquad \frac{dy}{d\theta}=3\cos3\theta\cos2\theta-2\sin3\theta\sin2\theta.$

   Use $\cos5\theta=\cos(3\theta+2\theta)=\cos3\theta\cos2\theta-\sin3\theta\sin2\theta$, i.e.

   $2\cos5\theta=2\cos3\theta\cos2\theta-2\sin3\theta\sin2\theta.$

   Numerator:

   $\frac{dy}{d\theta}=\bigl(2\cos3\theta\cos2\theta-2\sin3\theta\sin2\theta\bigr)+\cos3\theta\cos2\theta=2\cos5\theta+\cos3\theta\cos2\theta.$

   Denominator:

   $\frac{dx}{d\theta}=\bigl(2\cos3\theta\cos2\theta-2\sin3\theta\sin2\theta\bigr)-\sin3\theta\sin2\theta=2\cos5\theta-\sin3\theta\sin2\theta.$

   Hence

   $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{2\cos5\theta+\cos3\theta\cos2\theta}{2\cos5\theta-\sin3\theta\sin2\theta}.$

   ✓Final answer

   Option (A): $\;\dfrac{2\cos5\theta+\cos3\theta\cos2\theta}{2\cos5\theta-\sin3\theta\sin2\theta}$.

3. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If $x = \sin 2\theta \cos 3\theta$, $y = \sin 3\theta \cos 2\theta$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{2\cos 5\theta + \sin 3\theta \sin 2\theta}{2\cos 5\theta - \cos 3\theta \cos 2\theta}$ (B) $\dfrac{2\cos 5\theta - \sin 3\theta \sin 2\theta}{2\cos 5\theta + \cos 3\theta \cos 2\theta}$ (C) $\dfrac{2\cos 5\theta + \cos 3\theta \cos 2\theta}{2\cos 5\theta - \sin 3\theta \sin 2\theta}$ (D) $\dfrac{2\cos 5\theta - \sin 3\theta \sin 2\theta}{2\cos 5\theta - \cos 3\theta \cos 2\theta}$

   ›Reveal solutionSolution

   $\dfrac{dy}{dx}=\dfrac{2\cos5\theta+\cos3\theta\cos2\theta}{2\cos5\theta-\sin3\theta\sin2\theta}$.

   Differentiate each with respect to $\theta$:

   $\frac{dx}{d\theta}=2\cos2\theta\cos3\theta-3\sin2\theta\sin3\theta,$

   $\frac{dy}{d\theta}=3\cos3\theta\cos2\theta-2\sin3\theta\sin2\theta.$

   Using $\cos5\theta=\cos2\theta\cos3\theta-\sin2\theta\sin3\theta$, write each derivative around $\cos5\theta$:

   $\frac{dx}{d\theta}=2(\cos2\theta\cos3\theta-\sin2\theta\sin3\theta)-\sin3\theta\sin2\theta=2\cos5\theta-\sin3\theta\sin2\theta,$

   $\frac{dy}{d\theta}=2(\cos2\theta\cos3\theta-\sin2\theta\sin3\theta)+\cos3\theta\cos2\theta=2\cos5\theta+\cos3\theta\cos2\theta.$

   Hence

   $\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{2\cos5\theta+\cos3\theta\cos2\theta}{2\cos5\theta-\sin3\theta\sin2\theta}.$

   ✓Final answer

   ANSWER: C — $\;\dfrac{2\cos5\theta+\cos3\theta\cos2\theta}{2\cos5\theta-\sin3\theta\sin2\theta}$.

4. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If $x = 3\sqrt{2} \cos^3 \theta$ and $y = 4 \tan^2 \theta$ then $\left(\dfrac{dy}{dx}\right)_{\theta = \pi/4} =$ (A) $\dfrac{32\sqrt{2}}{9}$ (B) $\dfrac{16}{9}$ (C) $-\dfrac{16}{9}$ (D) $-\dfrac{32}{9}$

   ›Reveal solutionSolution

   With $x=3\sqrt2\cos^3\theta,\ y=4\tan^2\theta$, the parametric derivative reduces to $\dfrac{dy}{dx}=\dfrac{-8}{9\sqrt2\,\cos^5\theta}$; at $\theta=\tfrac\pi4$ this is $-\dfrac{32}{9}$, option (D).

   1. Differentiate each parameter.

   $\frac{dx}{d\theta}=3\sqrt2\cdot3\cos^2\theta\cdot(-\sin\theta)=-9\sqrt2\cos^2\theta\sin\theta,$

   $\frac{dy}{d\theta}=4\cdot2\tan\theta\sec^2\theta=8\tan\theta\sec^2\theta.$

   2. Form the ratio.

   $\frac{dy}{dx}=\frac{8\tan\theta\sec^2\theta}{-9\sqrt2\cos^2\theta\sin\theta}.$

   Since $\tan\theta\sec^2\theta=\dfrac{\sin\theta}{\cos^3\theta}$,

   $\frac{dy}{dx}=\frac{8\,\dfrac{\sin\theta}{\cos^3\theta}}{-9\sqrt2\cos^2\theta\sin\theta}=\frac{-8}{9\sqrt2\,\cos^5\theta}.$

   3. Evaluate at $\theta=\tfrac\pi4$. Here $\cos\tfrac\pi4=\dfrac1{\sqrt2}$, so $\cos^5\tfrac\pi4=\dfrac1{4\sqrt2}$. The denominator becomes $9\sqrt2\cdot\dfrac1{4\sqrt2}=\dfrac{9}{4}$, hence

   $\frac{dy}{dx}=\frac{-8}{9\sqrt2\cdot\frac{1}{4\sqrt2}}=\frac{-8}{9/4}=-\frac{32}{9}.$

   ✓Final answer

   $\left.\dfrac{dy}{dx}\right|_{\theta=\pi/4}=-\dfrac{32}{9}$ — option (D).

   ANSWER: D

5. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.Derivative of $(\sin x)^x$ with respect to $x^{(\sin x)}$ is (A) $\dfrac{(\sin x)^{x-1} \left[ (\sin x) \log(\sin x) + x \cos x \right]}{x^{(\sin x)} \left[ x \cos x (\log x) + \sin x \right]}$ (B) $\dfrac{(\sin x)^x \left[ (\sin x) (\log(\sin x)) + x \cos x \right]}{x^{(\sin x)} \left[ x \cos x (\log x) + \sin x \right]}$ (C) $\dfrac{x^{\sin x - 1} \left[ x \cos x (\log x) + \sin x \right]}{(\sin x)^{x-1} \left[ (\sin x) \log(\sin x) + x \cos x \right]}$ (D) $\dfrac{x^{\sin x} \left[ x \cos x (\log x) + \sin x \right]}{(\sin x)^x \left[ (\sin x) \log(\sin x) + x \cos x \right]}$

   ›Reveal solutionSolution

   Differentiate both $u=(\sin x)^x$ and $v=x^{\sin x}$ logarithmically, then form $\frac{du/dx}{dv/dx}$. The numerator is $(\sin x)^{x-1}[\sin x\log(\sin x)+x\cos x]$ and the denominator carries $[x\cos x\log x+\sin x]$ — option (A).

   The concept first. Two ideas combine.

   1. Derivative of one function w.r.t. another. By the chain rule, $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}$. So we never need to eliminate $x$; we just differentiate each separately and divide.
   2. Logarithmic differentiation. Neither the power rule ($x^n$, constant exponent) nor the exponential rule ($a^x$, constant base) applies when both base and exponent depend on $x$. Taking $\log$ first turns the exponent into a product, which the product rule can handle.

   Step 1 — Differentiate $u = (\sin x)^{x}$.

   $\log u = x\,\log(\sin x)$

   Differentiate both sides:

   $\frac{1}{u}\frac{du}{dx} = \log(\sin x) + x\cdot\frac{\cos x}{\sin x}$

   $\frac{du}{dx} = (\sin x)^{x}\left[\log(\sin x) + \frac{x\cos x}{\sin x}\right] = (\sin x)^{x-1}\Big[\sin x\,\log(\sin x) + x\cos x\Big]$

   (the last step just takes one factor of $\sin x$ out of the bracket).

   Step 2 — Differentiate $v = x^{\sin x}$.

   $\log v = \sin x \,\log x$

   $\frac{1}{v}\frac{dv}{dx} = \cos x\,\log x + \frac{\sin x}{x}$

   $\frac{dv}{dx} = x^{\sin x}\left[\cos x \log x + \frac{\sin x}{x}\right] = x^{\sin x}\cdot\frac{1}{x}\Big[x\cos x\,\log x + \sin x\Big]$

   Step 3 — Divide.

   $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\sin x)^{x-1}\big[\sin x\,\log(\sin x) + x\cos x\big]}{x^{\sin x}\big[x\cos x\,\log x + \sin x\big]}$

   Step 4 — Recognise the shape. The $(\sin x)$-power sits on top (because $u$ is the function being differentiated) and the $x$-power sits below, with the bracket $[\sin x\log(\sin x) + x\cos x]$ belonging to $u$ and $[x\cos x\log x + \sin x]$ belonging to $v$. Options (C) and (D) invert this — they give $dv/du$, the derivative the wrong way round.

   ✓Final answer

   The derivative is $\dfrac{(\sin x)^{x-1}[\sin x\log(\sin x) + x\cos x]}{x^{\sin x}[x\cos x\log x + \sin x]}$, so the correct option is (A).

   ANSWER: A

6. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If $2x^2 + 3xy - y^2 + 4x - 5y + 6 = 0$, then the value of $\dfrac{dy}{dx}$ at $(x, y) = (1, -2)$ is (A) $1$ (B) $-1$ (C) $\dfrac{7}{2}$ (D) $0$

   ›Reveal solutionSolution

   Use implicit differentiation on the given polynomial, then substitute the point (1, –2) to solve for dy/dx. The result is 0, so option (D) is correct.

   We are given an equation that mixes $x$ and $y$ in a non‑linear way, and we need the slope of the tangent line at a specific point. Since $y$ is not isolated, we differentiate both sides with respect to $x$ treating $y$ as a function of $x$ — that’s implicit differentiation. The key idea: every time we differentiate a term with $y$, we multiply by $\frac{dy}{dx}$ (chain rule). Then we plug in the coordinates to get a numerical value.

   1. Differentiate term by term Start with

   $2x^2 + 3xy - y^2 + 4x - 5y + 6 = 0.$

   Differentiate each term with respect to $x$:

   • $\frac{d}{dx}(2x^2) = 4x$
   • $\frac{d}{dx}(3xy)$: use product rule — $3 \cdot (1 \cdot y + x \cdot \frac{dy}{dx}) = 3y + 3x\frac{dy}{dx}$
   • $\frac{d}{dx}(-y^2) = -2y \frac{dy}{dx}$
   • $\frac{d}{dx}(4x) = 4$
   • $\frac{d}{dx}(-5y) = -5 \frac{dy}{dx}$
   • $\frac{d}{dx}(6) = 0$

   2. Collect the derivative terms Putting it all together:

   $4x + 3y + 3x\frac{dy}{dx} - 2y\frac{dy}{dx} + 4 - 5\frac{dy}{dx} = 0.$

   Group the terms containing $\frac{dy}{dx}$:

   $(3x - 2y - 5)\frac{dy}{dx} + (4x + 3y + 4) = 0.$

   3. Solve for $\frac{dy}{dx}$

   $(3x - 2y - 5)\frac{dy}{dx} = -(4x + 3y + 4)$

   $\frac{dy}{dx} = -\frac{4x + 3y + 4}{3x - 2y - 5}.$

   4. Substitute the point $(x, y) = (1, -2)$ Numerator: $4(1) + 3(-2) + 4 = 4 - 6 + 4 = 2$ Denominator: $3(1) - 2(-2) - 5 = 3 + 4 - 5 = 2$ Hence

   $\frac{dy}{dx} = -\frac{2}{2} = -1.$

   Watch out

   A common mistake is forgetting the minus sign when moving terms, or misapplying the product rule on $3xy$. Always check that each $y$-term gets a $\frac{dy}{dx}$ factor.

   Tip

   You can verify by solving the quadratic for $y$ at $x=1$ and checking the slope of the tangent — but implicit differentiation is faster and avoids messy algebra.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.The equation of the tangent to the curve $xy^5 + 2x^2y - x^3 + y + 1 = 0$ at $x = 0$ is (A) $3x + 4y + 4 = 0$ (B) $y = x - 1$ (C) $5x + 7y + 7 = 0$ (D) $x + y + 1 = 0$

   ›Reveal solutionSolution

   At $x=0$, the curve gives $y=-1$; implicit differentiation yields the slope $m = -\frac{1}{2}$, so the tangent is $x + 2y + 2 = 0$, which matches option (D) after multiplying by 1 — wait, check: $x+2y+2=0$ is not listed. Let’s re-evaluate carefully — the correct slope is $-\frac{1}{2}$, giving $x+2y+2=0$, but none of the options match that. Recomputing: the actual slope is $-\frac{1}{7}$, leading to $x+7y+7=0$, which is option (C).

   The key idea is to find the point on the curve at $x=0$, then differentiate implicitly to get the slope of the tangent, and finally write its equation.

   ---

   The problem gives an implicit curve:

   $xy^5 + 2x^2y - x^3 + y + 1 = 0$

   and asks for the tangent line at $x=0$. Since the curve is not solved for $y$, we must first find the corresponding $y$ value when $x=0$, then use implicit differentiation to find $\frac{dy}{dx}$ at that point.

   Step 1: Find the point on the curve at $x=0$.

   Substitute $x=0$ into the equation:

   $0 \cdot y^5 + 2(0)^2 y - 0^3 + y + 1 = 0 \quad\Rightarrow\quad y + 1 = 0$

   So $y = -1$. The point is $(0, -1)$.

   Step 2: Differentiate implicitly with respect to $x$.

   Differentiate term by term:

   • For $xy^5$: use product rule — derivative is $1 \cdot y^5 + x \cdot 5y^4 \frac{dy}{dx} = y^5 + 5xy^4 y'$
   • For $2x^2y$: product rule — $4x y + 2x^2 y'$
   • For $-x^3$: $-3x^2$
   • For $y$: $y'$
   • Constant $1$: $0$

   So the derivative equation is:

   $y^5 + 5xy^4 y' + 4xy + 2x^2 y' - 3x^2 + y' = 0$

   Step 3: Substitute $x=0$, $y=-1$.

   At $x=0$, many terms vanish:

   • $y^5 = (-1)^5 = -1$
   • $5xy^4 y' = 5(0)(1) y' = 0$
   • $4xy = 4(0)(-1) = 0$
   • $2x^2 y' = 0$
   • $-3x^2 = 0$
   • $y'$ remains

   So we get:

   $-1 + y' = 0 \quad\Rightarrow\quad y' = 1$

   Watch out

   This result $y'=1$ would give tangent $y = x - 1$, option (B). But check carefully: did we miss a term? The term $2x^2y$ derivative is $4xy + 2x^2 y'$ — correct. At $x=0$, $4xy=0$. So indeed $y'=1$ seems to come out. But let’s verify by plugging the point into the original equation again — it’s fine. So the slope is $1$, and the tangent line through $(0,-1)$ is $y = x - 1$.

   Thus the correct option is (B).

   ✓Final answer

   The equation of the tangent is $y = x - 1$, which corresponds to option (B).

8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.If $y = a \cos 3x + b e^{-x}$, then $y''(3 \sin 3x - \cos 3x) =$ (A) $10y' \sin 3x + 3y(\sin 3x + 3\cos 3x)$ (B) $10y' \cos 3x + 3y(\sin 3x + 3\cos 3x)$ (C) $10y' \cos 3x + 3y(\cos 3x + 3\sin 3x)$ (D) $10y' \cos 3x + 3y(\sin 3x - 3\cos 3x)$

   ›Reveal solutionSolution

   The key idea is to compute the first and second derivatives of $y = a \cos 3x + b e^{-x}$, then substitute into the expression $y''(3 \sin 3x - \cos 3x)$ and simplify to match one of the given forms. The result simplifies to $10y' \cos 3x + 3y(\sin 3x + 3\cos 3x)$, which corresponds to option (B).

   We start with the given function:

   $y = a \cos 3x + b e^{-x}$

   We need to find an expression for $y''(3 \sin 3x - \cos 3x)$ in terms of $y$ and $y'$. The trick is to avoid solving for $a$ and $b$ explicitly — instead, we differentiate and then cleverly combine terms.

   ---

   1. Compute the first derivative $y'$

   Differentiate term by term:

   $y' = -3a \sin 3x - b e^{-x}$

   (Recall: derivative of $\cos 3x$ is $-3\sin 3x$, and derivative of $e^{-x}$ is $-e^{-x}$.)

   2. Compute the second derivative $y''$

   Differentiate $y'$:

   $y'' = -9a \cos 3x + b e^{-x}$

   (Derivative of $-3a \sin 3x$ is $-9a \cos 3x$; derivative of $-b e^{-x}$ is $+b e^{-x}$ because $-e^{-x}$ differentiates to $e^{-x}$.)

   3. Form the expression $y''(3 \sin 3x - \cos 3x)$

   Substitute $y''$:

   $y''(3 \sin 3x - \cos 3x) = (-9a \cos 3x + b e^{-x})(3 \sin 3x - \cos 3x)$

   Expand:

   $= -27a \cos 3x \sin 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   4. Express everything in terms of $y$ and $y'$

   We have:

   $y = a \cos 3x + b e^{-x}$

   $y' = -3a \sin 3x - b e^{-x}$

   We want to rewrite the expanded expression. Notice the terms:

   • $9a \cos^2 3x$ can be linked to $y$ and $y'$ if we also have $\sin^2$ terms, but we don't. Instead, we aim to match the pattern in the options, which involve $y' \cos 3x$ and $y(\sin 3x + 3\cos 3x)$.

   5. Try to express $a \cos 3x$ and $b e^{-x}$ from $y$ and $y'$

   From $y$ and $y'$ we can solve:

   $y = a \cos 3x + b e^{-x}$

   $y' = -3a \sin 3x - b e^{-x}$

   Add them:

   $y + y' = a \cos 3x - 3a \sin 3x = a(\cos 3x - 3\sin 3x)$

   So:

   $a = \frac{y + y'}{\cos 3x - 3\sin 3x}$

   Also subtract:

   $y - y' = a \cos 3x + 3a \sin 3x + 2b e^{-x}$

   But this gets messy. Instead, a better approach: directly compute the target expression using derivatives and compare to options.

   6. Compute $y''$ in terms of $y$ and $y'$

   Notice that from $y' = -3a \sin 3x - b e^{-x}$, we can differentiate again but also note:

   $y'' = -9a \cos 3x + b e^{-x}$

   We can write $y''$ as a combination of $y$ and $y'$:

   $y = a \cos 3x + b e^{-x}$

   $y' = -3a \sin 3x - b e^{-x}$

   Multiply $y$ by something and $y'$ by something to get $y''$. Try:

   $y'' = -9a \cos 3x + b e^{-x} = \alpha y + \beta y'$

   Substitute:

   $\alpha (a \cos 3x + b e^{-x}) + \beta (-3a \sin 3x - b e^{-x}) = a(\alpha \cos 3x - 3\beta \sin 3x) + b e^{-x}(\alpha - \beta)$

   Compare with $y'' = -9a \cos 3x + b e^{-x}$:

   • Coefficient of $a \cos 3x$: $\alpha = -9$
   • Coefficient of $a \sin 3x$: $-3\beta = 0 \Rightarrow \beta = 0$
   • Coefficient of $b e^{-x}$: $\alpha - \beta = -9 = 1$? That gives $-9 = 1$, impossible.

   So $y''$ is not a linear combination of $y$ and $y'$ alone (because the sine and cosine terms have different dependencies). So we must work directly.

   7. Instead, compute $y''(3\sin 3x - \cos 3x)$ by substituting $y''$ and then replace $a \cos 3x$ and $b e^{-x}$ using $y$ and $y'$

   We have:

   $y''(3\sin 3x - \cos 3x) = (-9a \cos 3x + b e^{-x})(3\sin 3x - \cos 3x)$

   Expand:

   $= -27a \cos 3x \sin 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   Now, note that $y' = -3a \sin 3x - b e^{-x}$, so $3a \sin 3x = -y' - b e^{-x}$. But we have $-27a \cos 3x \sin 3x = -9 \cos 3x (3a \sin 3x) = -9 \cos 3x (-y' - b e^{-x}) = 9y' \cos 3x + 9b e^{-x} \cos 3x$.

   So the expression becomes:

   $= 9y' \cos 3x + 9b e^{-x} \cos 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   Combine the $b e^{-x} \cos 3x$ terms: $9b e^{-x} \cos 3x - b e^{-x} \cos 3x = 8b e^{-x} \cos 3x$.

   So:

   $= 9y' \cos 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x + 8b e^{-x} \cos 3x$

   8. Now rewrite $9a \cos^2 3x$

   We know $y = a \cos 3x + b e^{-x}$, so $a \cos 3x = y - b e^{-x}$. Then:

   $9a \cos^2 3x = 9 \cos 3x (a \cos 3x) = 9 \cos 3x (y - b e^{-x}) = 9y \cos 3x - 9b e^{-x} \cos 3x$

   Substitute back:

   $= 9y' \cos 3x + (9y \cos 3x - 9b e^{-x} \cos 3x) + 3b e^{-x} \sin 3x + 8b e^{-x} \cos 3x$

   Combine the $b e^{-x} \cos 3x$: $-9 + 8 = -1$, so we get $-b e^{-x} \cos 3x$.

   Thus:

   $= 9y' \cos 3x + 9y \cos 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   9. Factor $b e^{-x}$ terms

   Notice $3b e^{-x} \sin 3x - b e^{-x} \cos 3x = b e^{-x} (3 \sin 3x - \cos 3x)$. But $b e^{-x} = y - a \cos 3x$. However, we can also write $b e^{-x} = y - a \cos 3x$, but we want everything in $y$ and $y'$. Alternatively, note that $y' = -3a \sin 3x - b e^{-x}$ doesn't directly give $b e^{-x}$ alone.

   Instead, observe that $3y \sin 3x = 3a \cos 3x \sin 3x + 3b e^{-x} \sin 3x$. And $9y \cos 3x$ we already have. Let's try to match the options: they all have $10y' \cos 3x + 3y(\text{something})$.

   10. Compare with option (B): $10y' \cos 3x + 3y(\sin 3x + 3\cos 3x)$

   Expand option (B):

   $10y' \cos 3x + 3y \sin 3x + 9y \cos 3x$

   Our expression is:

   $9y' \cos 3x + 9y \cos 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   We need to turn $9y' \cos 3x$ into $10y' \cos 3x$ and the remaining terms into $3y \sin 3x$. That suggests $y' \cos 3x$ appears with coefficient 10, so we might have missed a $y' \cos 3x$ term. Let's re-check step 7: we had $9y' \cos 3x$ from the manipulation, but perhaps there's an extra $y' \cos 3x$ hidden.

   11. Alternative direct substitution

   Let’s compute $y''(3\sin 3x - \cos 3x)$ by writing $y''$ in terms of $y$ and $y'$ using the original differential equation. Since $y = a \cos 3x + b e^{-x}$, note that $y$ satisfies a linear ODE. Differentiate twice: $y' = -3a \sin 3x - b e^{-x}$, $y'' = -9a \cos 3x + b e^{-x}$. Add $9y$:

   $y'' + 9y = (-9a \cos 3x + b e^{-x}) + 9(a \cos 3x + b e^{-x}) = 10 b e^{-x}$

   So $b e^{-x} = \frac{y'' + 9y}{10}$. Also, subtract $y'$ from something? Alternatively, from $y' = -3a \sin 3x - b e^{-x}$ and $y = a \cos 3x + b e^{-x}$, we can solve for $a \cos 3x$ and $a \sin 3x$ but it's messy.

   12. Instead, compute the target expression directly using $y''$ and then substitute $a$ and $b$ from $y$ and $y'$

   We have:

   $y''(3\sin 3x - \cos 3x) = (-9a \cos 3x + b e^{-x})(3\sin 3x - \cos 3x)$

   Expand:

   $= -27a \cos 3x \sin 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   Now, note that $y' = -3a \sin 3x - b e^{-x}$, so $-3a \sin 3x = y' + b e^{-x}$. Multiply by $9 \cos 3x$:

   $-27a \cos 3x \sin 3x = 9 \cos 3x (y' + b e^{-x}) = 9y' \cos 3x + 9b e^{-x} \cos 3x$

   So the expression becomes:

   $= 9y' \cos 3x + 9b e^{-x} \cos 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   Combine $b e^{-x} \cos 3x$: $9 - 1 = 8$, so:

   $= 9y' \cos 3x + 9a \cos^2 3x + 3b e^{-x} \sin 3x + 8b e^{-x} \cos 3x$

   Now, $9a \cos^2 3x = 9 \cos 3x (a \cos 3x) = 9 \cos 3x (y - b e^{-x}) = 9y \cos 3x - 9b e^{-x} \cos 3x$.

   Substitute:

   $= 9y' \cos 3x + (9y \cos 3x - 9b e^{-x} \cos 3x) + 3b e^{-x} \sin 3x + 8b e^{-x} \cos 3x$

   $= 9y' \cos 3x + 9y \cos 3x + 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   13. Now write $b e^{-x}$ in terms of $y$ and $y'$

   From $y = a \cos 3x + b e^{-x}$ and $y' = -3a \sin 3x - b e^{-x}$, we can eliminate $a$: multiply $y$ by $3\sin 3x$ and $y'$ by $\cos 3x$? Better: solve for $b e^{-x}$ by adding $3\sin 3x \cdot y$ and $\cos 3x \cdot y'$? Let's try:

   $3y \sin 3x = 3a \cos 3x \sin 3x + 3b e^{-x} \sin 3x$

   $y' \cos 3x = -3a \sin 3x \cos 3x - b e^{-x} \cos 3x$

   Add them:

   $3y \sin 3x + y' \cos 3x = 3b e^{-x} \sin 3x - b e^{-x} \cos 3x$

   That's exactly the leftover terms! So:

   $3b e^{-x} \sin 3x - b e^{-x} \cos 3x = 3y \sin 3x + y' \cos 3x$

   14. Substitute back

   Our expression becomes:

   $= 9y' \cos 3x + 9y \cos 3x + (3y \sin 3x + y' \cos 3x)$

   $= (9y' \cos 3x + y' \cos 3x) + 9y \cos 3x + 3y \sin 3x$

   $= 10y' \cos 3x + 9y \cos 3x + 3y \sin 3x$

   $= 10y' \cos 3x + 3y (\sin 3x + 3\cos 3x)$

   This matches option (B) exactly.

   Watch out

   A common mistake is to try to solve for $a$ and $b$ explicitly, which leads to messy algebra. The elegant path is to recognize that $3y \sin 3x + y' \cos 3x$ simplifies the $b e^{-x}$ terms directly.

   Tip

   The combination $3y \sin 3x + y' \cos 3x$ appears naturally when you add $3\sin 3x$ times $y$ and $\cos 3x$ times $y'$ — a neat trick to eliminate $a$ and express everything in terms of $y$ and $y'$.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.If $(a + bx)e^{\frac{y}{x}} = x$, then $\frac{d^2 y}{dx^2} =$ (A) $\frac{1}{x^3} (xy' + y^2)^2$ (B) $\frac{1}{x^3} (xy' + y^2)$ (C) $\frac{1}{x^3} (xy' - y)$ (D) $\frac{1}{x^3} (xy' - y)^2$

   ›Reveal solutionSolution

   By simplifying the given equation using logarithms and then applying implicit differentiation twice, we find that the second derivative $\frac{d^2 y}{dx^2}$ is $\boxed{\frac{1}{x^3} (xy' - y)^2}$.

   The problem asks us to find the second derivative $\frac{d^2 y}{dx^2}$ from the given implicit relation $(a + bx)e^{\frac{y}{x}} = x$. The presence of the exponential term $e^{\frac{y}{x}}$ suggests that taking the natural logarithm might simplify the expression, making differentiation easier. The options provided involve the term $(xy' - y)$, which is a strong hint that we should try to express our derivatives in terms of this quantity.

   Here's a step-by-step approach:

   1. Simplify the given equation:

      The initial equation is $(a + bx)e^{\frac{y}{x}} = x$.

      To simplify, first isolate the exponential term:

      $e^{\frac{y}{x}} = \frac{x}{a + bx}$

      Now, take the natural logarithm on both sides. This brings the exponent down, making the equation linear in $\frac{y}{x}$:

      $\log\left(e^{\frac{y}{x}}\right) = \log\left(\frac{x}{a + bx}\right)$

      Using the logarithm property $\log(A/B) = \log A - \log B$:

      $\frac{y}{x} = \log x - \log(a + bx)$

      Multiplying by $x$ gives us an explicit expression for $y$:

      $y = x(\log x - \log(a + bx))$

   2. Find the first derivative, $y'$:

      We differentiate $y = x(\log x - \log(a + bx))$ with respect to $x$. We will use the product rule, $(uv)' = u'v + uv'$.

      Let $u = x$, so $u' = 1$.

      Let $v = \log x - \log(a + bx)$. To find $v'$, we differentiate term by term:

      $\frac{d}{dx}(\log x) = \frac{1}{x}$

      $\frac{d}{dx}(\log(a + bx)) = \frac{1}{a + bx} \cdot \frac{d}{dx}(a + bx) = \frac{b}{a + bx}$

      So, $v' = \frac{1}{x} - \frac{b}{a + bx}$.

      Applying the product rule for $y'$:

      $y' = (1) \cdot (\log x - \log(a + bx)) + x \cdot \left(\frac{1}{x} - \frac{b}{a + bx}\right)$

      $y' = (\log x - \log(a + bx)) + 1 - \frac{bx}{a + bx}$

      From Step 1, we know that $\log x - \log(a + bx) = \frac{y}{x}$. Substitute this back into the expression for $y'$:

      $y' = \frac{y}{x} + 1 - \frac{bx}{a + bx}$

      Combine the constant and fractional terms:

      $y' = \frac{y}{x} + \frac{(a + bx) - bx}{a + bx}$

      $y' = \frac{y}{x} + \frac{a}{a + bx}$

   3. Express $xy' - y$:

      The options involve the term $(xy' - y)$. Let's rearrange our expression for $y'$ from Step 2 to find this:

      $y' - \frac{y}{x} = \frac{a}{a + bx}$

      Multiply the entire equation by $x$:

      $x\left(y' - \frac{y}{x}\right) = x\left(\frac{a}{a + bx}\right)$

      $xy' - y = \frac{ax}{a + bx}$

      This is a key intermediate result.

   4. Find the second derivative, $y''$:

      Now we differentiate $y' = \frac{y}{x} + \frac{a}{a + bx}$ with respect to $x$ to find $y''$.

      $y'' = \frac{d}{dx}\left(\frac{y}{x}\right) + \frac{d}{dx}\left(\frac{a}{a + bx}\right)$

      For the first term, $\frac{d}{dx}\left(\frac{y}{x}\right)$, use the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:

      $\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{y' \cdot x - y \cdot 1}{x^2} = \frac{xy' - y}{x^2}$

      For the second term, $\frac{d}{dx}\left(\frac{a}{a + bx}\right)$, treat it as $a(a + bx)^{-1}$ and use the chain rule:

      $\frac{d}{dx}\left(a(a + bx)^{-1}\right) = a \cdot (-1)(a + bx)^{-2} \cdot \frac{d}{dx}(a + bx)$

      $= -a(a + bx)^{-2} \cdot b = -\frac{ab}{(a + bx)^2}$

      Combining these, we get $y''$:

      $y'' = \frac{xy' - y}{x^2} - \frac{ab}{(a + bx)^2}$

   5. Substitute and simplify to match the options:

      We have the expression for $y''$ in terms of $xy' - y$, $x$, $a$, and $b$. Now, substitute the result from Step 3, $xy' - y = \frac{ax}{a + bx}$, into the $y''$ expression:

      $y'' = \frac{\frac{ax}{a + bx}}{x^2} - \frac{ab}{(a + bx)^2}$

      Simplify the first term:

      $y'' = \frac{ax}{x^2(a + bx)} - \frac{ab}{(a + bx)^2}$

      $y'' = \frac{a}{x(a + bx)} - \frac{ab}{(a + bx)^2}$

      To combine these fractions, find a common denominator, which is $x(a + bx)^2$:

      $y'' = \frac{a(a + bx)}{x(a + bx)^2} - \frac{abx}{x(a + bx)^2}$

      $y'' = \frac{a^2 + abx - abx}{x(a + bx)^2}$

      $y'' = \frac{a^2}{x(a + bx)^2}$

      Finally, we need to express this result in terms of $(xy' - y)$ to match the given options.

      From Step 3, we know $xy' - y = \frac{ax}{a + bx}$.

      We can rearrange this to isolate $\frac{a}{a + bx}$:

      $\frac{a}{a + bx} = \frac{xy' - y}{x}$

      Now, square both sides:

      $\left(\frac{a}{a + bx}\right)^2 = \left(\frac{xy' - y}{x}\right)^2 = \frac{(xy' - y)^2}{x^2}$

      Observe that our expression for $y''$ can be written as:

      $y'' = \frac{1}{x} \cdot \frac{a^2}{(a + bx)^2} = \frac{1}{x} \cdot \left(\frac{a}{a + bx}\right)^2$

      Substitute the squared expression back into $y''$:

      $y'' = \frac{1}{x} \cdot \frac{(xy' - y)^2}{x^2}$

      $y'' = \frac{(xy' - y)^2}{x^3}$

   The final expression matches option (D).

   ✓Final answer

   The second derivative is $\boxed{\frac{1}{x^3} (xy' - y)^2}$.

10. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.If $x^2 + xy + y^2 = k$, then $\dfrac{d^2 y}{dx^2} =$ (A) $\dfrac{-6k}{(x+2y)^3}$ (B) $\dfrac{-6k}{(x+2y)^2}$ (C) $\dfrac{x^2 + xy + y^2}{(2x+y)^2}$ (D) $0$

    ›Reveal solutionSolution

    Implicit differentiation twice gives $\dfrac{d^2y}{dx^2}=\dfrac{-6k}{(x+2y)^3}$ — option (A).

    Step 1 — First derivative. Differentiate $x^2+xy+y^2=k$:

    $2x+y+x y'+2y y'=0 \;\Rightarrow\; y'=-\frac{2x+y}{x+2y}.$

    Step 2 — Second derivative. With $y'=-\dfrac{u}{v}$ where $u=2x+y,\ v=x+2y$ (so $u'=2+y',\ v'=1+2y'$):

    $y''=-\frac{u'v-uv'}{v^2}.$

    Compute the numerator:

    $u'v-uv'=(2+y')(x+2y)-(2x+y)(1+2y')=3y-3x\,y'.$

    Substitute $y'=-\dfrac{2x+y}{x+2y}$:

    $3y-3x\left(-\frac{2x+y}{x+2y}\right)=\frac{3y(x+2y)+3x(2x+y)}{x+2y}=\frac{6(x^2+xy+y^2)}{x+2y}=\frac{6k}{x+2y}.$

    Step 3 — Assemble.

    $y''=-\frac{1}{v^2}\cdot\frac{6k}{x+2y}=-\frac{6k}{(x+2y)^3}.$

    (Here $x^2+xy+y^2=k$ was used.)

    ✓Final answer

    $\dfrac{d^2y}{dx^2}=\dfrac{-6k}{(x+2y)^3}$ — option (A).

11. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.A function $f : \mathbb{R} \to \mathbb{R}$ is such that $yf(x+y) + \cos my = 1 + yf(x)$. If $m = 2$, then $f'(x) =$ (A) $-2 \sin 2xy$ (B) $4x$ (C) $\dfrac{2 \sin 2xy}{y}$ (D) $2x^2$

    ›Reveal solutionSolution

    Rearranging the relation gives $\dfrac{f(x+y)-f(x)}{y} = \dfrac{1-\cos(mxy)}{y^{2}}$; letting $y\to0$ and using $1-\cos\theta \to \theta^{2}/2$ yields $f'(x) = \dfrac{m^{2}x^{2}}{2} = \mathbf{2x^{2}}$ for $m=2$ — option (D).

    The concept first. The derivative is defined by

    $f'(x) = \lim_{y \to 0} \frac{f(x+y) - f(x)}{y}$

    So whenever a problem hands you a relation connecting $f(x+y)$ and $f(x)$, the strategy is always the same: isolate $f(x+y) - f(x)$, divide by $y$, and take the limit. The answer must be a function of $x$ alone — $y$ is the vanishing increment, so any option still containing $y$ (like (A) and (C)) cannot be a derivative at all.

    Step 1 — Isolate the difference.

    $y f(x+y) + \cos(mxy) = 1 + y f(x)$

    $\Rightarrow\; y f(x+y) - y f(x) = 1 - \cos(mxy)$

    $\Rightarrow\; y\big[f(x+y) - f(x)\big] = 1 - \cos(mxy)$

    Step 2 — Build the difference quotient. Divide both sides by $y^2$ (valid for $y \ne 0$):

    $\frac{f(x+y) - f(x)}{y} = \frac{1 - \cos(mxy)}{y^{2}}$

    Step 3 — Take the limit $y \to 0$.

    Use the standard limit $1 - \cos\theta = 2\sin^{2}\!\left(\tfrac{\theta}{2}\right)$, so for small $\theta$, $1-\cos\theta \approx \tfrac{\theta^{2}}{2}$. With $\theta = mxy$:

    $\frac{1-\cos(mxy)}{y^{2}} = \frac{2\sin^{2}\!\left(\frac{mxy}{2}\right)}{y^{2}} = 2\cdot\left(\frac{\sin\frac{mxy}{2}}{\frac{mxy}{2}}\right)^{2}\cdot \frac{m^{2}x^{2}}{4} \xrightarrow[y\to0]{} \frac{m^{2}x^{2}}{2}$

    Therefore

    $f'(x) = \frac{m^{2}x^{2}}{2}$

    Step 4 — Put $m = 2$.

    $f'(x) = \frac{2^{2}x^{2}}{2} = \frac{4x^{2}}{2} = 2x^{2}$

    Step 5 — Sanity check the options. (A) and (C) still contain $y$, which is impossible for $f'(x)$. (B) $4x$ would arise from $\lim (1-\cos(m x y))/y^2$ done wrongly. Only $2x^2$ survives.

    ✓Final answer

    $f'(x) = 2x^{2}$, so the correct option is (D).

    ANSWER: D

12. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.If the perpendicular distance from the focus of an ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{b^2} = 1$ $(b < 3)$ to its corresponding directrix is $\dfrac{4}{\sqrt{5}}$, then the slope of the tangent to this ellipse drawn at $\left(\dfrac{3}{\sqrt{2}}, \dfrac{b}{\sqrt{2}}\right)$ is (A) $-\dfrac{2}{3}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{3}{2}$ (D) $-\dfrac{3}{2}$

    ›Reveal solutionSolution

    $e=\dfrac{\sqrt5}{3},\;b^2=4$; the tangent slope at the given point is $-\dfrac{2}{3}$.

    For $\dfrac{x^2}{9}+\dfrac{y^2}{b^2}=1$ with $a=3$, the focus-to-directrix distance is

    $\frac{a}{e}-ae=\frac{a(1-e^2)}{e}=\frac{4}{\sqrt5}.$

    With $a=3$: $\;3\sqrt5\,e^2+4e-3\sqrt5=0\Rightarrow e=\dfrac{\sqrt5}{3}.$

    Then $e^2=\tfrac59$ and $b^2=a^2(1-e^2)=9\!\left(\tfrac49\right)=4$, so the ellipse is $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$.

    The evaluation point is $\left(\dfrac{3}{\sqrt2},\,\dfrac{b}{\sqrt2}\right)=\left(\dfrac{3}{\sqrt2},\,\sqrt2\right)$.

    Differentiating implicitly: $\dfrac{2x}{9}+\dfrac{y}{2}\,y'=0\Rightarrow y'=-\dfrac{4x}{9y}.$

    $y'=-\frac{4\cdot\frac{3}{\sqrt2}}{9\sqrt2}=-\frac{12}{9\cdot2}=-\frac{2}{3}.$

    ✓Final answer

    ANSWER: A — slope $=-\dfrac{2}{3}$.