Q.Differentiate the following w.r.t. $x$: $e^{x^3}$

Showing the 12 most recent of 26 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $f(x)$ is a differentiable function and $y = e^{f(x)} + e^{f(x)} + e^{f(x)} + \ldots \infty$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{y f'(x)}{1 + y}$ (B) $\dfrac{(1 + y) f'(x)}{y}$ (C) $\dfrac{(1 - y) f'(x)}{y}$ (D) $\dfrac{y f'(x)}{1 - y}$

   ›Reveal solutionSolution

   The expression is an infinite nested exponent, so it satisfies $y = e^{\,f(x)+y}$. Implicit differentiation gives $\dfrac{dy}{dx} = \dfrac{y\,f'(x)}{1-y}$.

   Setting up the self-similar equation. The right-hand side is an infinitely nested tower $y = e^{\,f(x)+e^{\,f(x)+\cdots}}$. Because the exponent contains an exact copy of the whole expression, the tower folds into itself:

   $y = e^{\,f(x) + y}$

   Take logarithms:

   $\log y = f(x) + y$

   Differentiate both sides with respect to $x$:

   $\frac{1}{y}\frac{dy}{dx} = f'(x) + \frac{dy}{dx}$

   Collect the derivative terms:

   $\frac{1}{y}\frac{dy}{dx} - \frac{dy}{dx} = f'(x) \;\Longrightarrow\; \frac{dy}{dx}\left(\frac{1-y}{y}\right) = f'(x)$

   Solve for the derivative:

   $\frac{dy}{dx} = \frac{y\,f'(x)}{1-y}$

   Note: as printed, the stem shows the exponent written additively; the standard reading of such "$\ldots\infty$" tower problems is the nested exponential $y=e^{f(x)+y}$, which the given options confirm.

   ✓Final answer

   $\dfrac{dy}{dx} = \dfrac{y\,f'(x)}{1-y}$, so the correct option is (D).

2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $f:\mathbb{R}-\{0\} \to \mathbb{R}$ is a differentiable function such that $\frac{1}{3}f(x) + 3f\left(\frac{1}{x}\right) = x - \frac{10}{3}$, then $f'(3) - f'\left(\frac{1}{3}\right) =$ (A) $\frac{12}{5}$ (B) $\frac{80}{9}$ (C) $3$ (D) $5$

   ›Reveal solutionSolution

   We differentiate the given functional equation with respect to $x$ and then substitute $x=3$ to directly find the required expression. The value of $f'(3) - f'\left(\frac{1}{3}\right)$ is $\boxed{3}$.

   The problem presents a functional equation involving $f(x)$ and $f(1/x)$, and asks for an expression involving their derivatives, $f'(3)$ and $f'(1/3)$. The most direct approach to solve such problems is to differentiate the given functional equation.

   Here's why this approach works:

   When you have an equation relating $f(x)$ and $f(1/x)$, differentiating it will introduce $f'(x)$ and $f'(1/x)$. The chain rule will be crucial for the term $f(1/x)$. After differentiation, we will have a new equation involving derivatives. By carefully choosing a value for $x$ (in this case, $x=3$), we can make the arguments of the derivatives match the terms we need to find.

   Let's work through the steps.

   1. Write down the given functional equation: We are given the equation:

   $\frac{1}{3}f(x) + 3f\left(\frac{1}{x}\right) = x - \frac{10}{3}$

   This equation holds for all $x \in \mathbb{R}-\{0\}$.
   

   2. Differentiate both sides with respect to $x$:

   Since the function $f(x)$ is differentiable, we can differentiate both sides of the equation with respect to $x$.

   Recall the chain rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$.

   Here, for the term $f(1/x)$, $g(x) = 1/x$, so $g'(x) = -1/x^2$.

   Differentiating the left side:
   

   $\frac{d}{dx}\left[\frac{1}{3}f(x) + 3f\left(\frac{1}{x}\right)\right] = \frac{1}{3}f'(x) + 3f'\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$

   $= \frac{1}{3}f'(x) - \frac{3}{x^2}f'\left(\frac{1}{x}\right)$

   Differentiating the right side:
   

   $\frac{d}{dx}\left[x - \frac{10}{3}\right] = 1 - 0 = 1$

   Equating the derivatives of both sides, we get:
   

   $\frac{1}{3}f'(x) - \frac{3}{x^2}f'\left(\frac{1}{x}\right) = 1$

   This is a new functional equation involving the derivatives.
   

   3. Substitute $x=3$ into the differentiated equation:

   We need to find $f'(3) - f'(1/3)$. Notice that if we substitute $x=3$ into the equation from Step 2, the arguments of $f'$ will become $3$ and $1/3$, which is exactly what we need.

   Substitute $x=3$:

   $\frac{1}{3}f'(3) - \frac{3}{(3)^2}f'\left(\frac{1}{3}\right) = 1$

   $\frac{1}{3}f'(3) - \frac{3}{9}f'\left(\frac{1}{3}\right) = 1$

   $\frac{1}{3}f'(3) - \frac{1}{3}f'\left(\frac{1}{3}\right) = 1$

   4. Simplify the expression: To isolate $f'(3) - f'(1/3)$, multiply the entire equation by $3$:

   $3 \left( \frac{1}{3}f'(3) - \frac{1}{3}f'\left(\frac{1}{3}\right) \right) = 3 \cdot 1$

   $f'(3) - f'\left(\frac{1}{3}\right) = 3$

   This directly gives us the value we were asked to find.

   ✓Final answer

   The value of $f'(3) - f'\left(\frac{1}{3}\right)$ is $\boxed{3}$.

3. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If $y = \tan^{-1} \left[ \dfrac{\sin^3(2x) - 3x^2 \sin(2x)}{3x \sin^2(2x) - x^3} \right]$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{6x \cos(2x) - 3 \sin(2x)}{x^2 - \sin^2(2x)}$ (B) $\dfrac{6x \sin(2x) - 3 \cos(2x)}{x^2 + \sin^2(2x)}$ (C) $\dfrac{2x \cos(2x) - \sin(2x)}{x^2 + \sin^2(2x)}$ (D) $\dfrac{6x \cos(2x) - 3 \sin(2x)}{x^2 + \sin^2(2x)}$

   ›Reveal solutionSolution

   The key is to recognise the argument of $\tan^{-1}$ as the tangent triple-angle formula $\tan(3\theta)$ with $\theta = \tan^{-1}\!\left(\frac{\sin(2x)}{x}\right)$, so $y = 3\tan^{-1}\!\left(\frac{\sin(2x)}{x}\right)$; differentiating gives $\frac{dy}{dx} = \frac{6x\cos(2x)-3\sin(2x)}{x^2+\sin^2(2x)}$, which matches option (D).

   ---

   The expression inside the inverse tangent looks messy — a ratio of two cubic-looking polynomials in $\sin(2x)$ and $x$. That structure is a dead giveaway for the triple-angle formula for tangent:

   $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$

   But here we have $\frac{\sin^3(2x) - 3x^2\sin(2x)}{3x\sin^2(2x) - x^3}$. If we set $\tan\theta = \frac{\sin(2x)}{x}$, then:

   • Numerator: $\sin^3(2x) - 3x^2\sin(2x) = x^3\left[\left(\frac{\sin(2x)}{x}\right)^3 - 3\left(\frac{\sin(2x)}{x}\right)\right] = x^3(\tan^3\theta - 3\tan\theta)$
   • Denominator: $3x\sin^2(2x) - x^3 = x^3\left[3\left(\frac{\sin(2x)}{x}\right)^2 - 1\right] = x^3(3\tan^2\theta - 1)$

   So the fraction becomes:

   $\frac{x^3(\tan^3\theta - 3\tan\theta)}{x^3(3\tan^2\theta - 1)} = \frac{\tan^3\theta - 3\tan\theta}{3\tan^2\theta - 1}$

   But $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = -\frac{\tan^3\theta - 3\tan\theta}{3\tan^2\theta - 1}$. So our fraction is actually $-\tan(3\theta)$. However, $\tan^{-1}(-\tan(3\theta)) = -3\theta$ (for appropriate principal values). Thus:

   $y = \tan^{-1}\!\left[-\tan(3\theta)\right] = -3\theta = -3\tan^{-1}\!\left(\frac{\sin(2x)}{x}\right)$

   Now differentiate.

   ---

   1. Differentiate $y = -3\tan^{-1}(u)$ where $u = \frac{\sin(2x)}{x}$.

   $\frac{dy}{dx} = -3 \cdot \frac{1}{1+u^2} \cdot \frac{du}{dx}$

   2. Find $\frac{du}{dx}$ using the quotient rule:

   $u = \frac{\sin(2x)}{x} \quad\Rightarrow\quad \frac{du}{dx} = \frac{2x\cos(2x) - \sin(2x)}{x^2}$

   3. Compute $1+u^2$:

   $1+u^2 = 1 + \frac{\sin^2(2x)}{x^2} = \frac{x^2 + \sin^2(2x)}{x^2}$

   4. Put it together:

   $\frac{dy}{dx} = -3 \cdot \frac{1}{\frac{x^2+\sin^2(2x)}{x^2}} \cdot \frac{2x\cos(2x) - \sin(2x)}{x^2} = -3 \cdot \frac{x^2}{x^2+\sin^2(2x)} \cdot \frac{2x\cos(2x) - \sin(2x)}{x^2}$

   The $x^2$ cancels, giving:

   $\frac{dy}{dx} = -3 \cdot \frac{2x\cos(2x) - \sin(2x)}{x^2+\sin^2(2x)} = \frac{-6x\cos(2x) + 3\sin(2x)}{x^2+\sin^2(2x)}$

   5. Multiply numerator and denominator by $-1$ to match the options:

   $\frac{dy}{dx} = \frac{6x\cos(2x) - 3\sin(2x)}{x^2+\sin^2(2x)}$

   Tip

   The triple-angle trick works because the coefficients (1 and 3) appear symmetrically. Always check if a complicated $\tan^{-1}$ argument matches $\tan(3\theta)$ or $\tanh$ identities — it saves pages of differentiation.

   Watch out

   A common mistake is forgetting the negative sign from $\tan(3\theta) = -\frac{\tan^3\theta - 3\tan\theta}{3\tan^2\theta - 1}$. If you miss it, you’ll get the wrong sign in the final derivative.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $x = \log p$ and $y = \dfrac{1}{p}$ then $\dfrac{dy}{dx} =$ (A) $-e^{-x}$ (B) $e^{-x}$ (C) $x$ (D) $y$

   ›Reveal solutionSolution

   Express $y$ as a function of $x$ (namely $y = e^{-x}$) and differentiate.

   Concept. When two variables are given in terms of a common parameter, eliminate the parameter (or use $\frac{dy}{dx} = \frac{dy/dp}{dx/dp}$).

   Step 1 — eliminate $p$. From $x = \log p$ we get $p = e^x$. Hence

   $y = \frac{1}{p} = e^{-x}.$

   Step 2 — differentiate.

   $\frac{dy}{dx} = \frac{d}{dx}\left(e^{-x}\right) = -e^{-x}.$

   Check by parametric differentiation. $\dfrac{dy}{dp} = -\dfrac{1}{p^2}$, $\dfrac{dx}{dp} = \dfrac{1}{p}$, so $\dfrac{dy}{dx} = \dfrac{-1/p^2}{1/p} = -\dfrac{1}{p} = -e^{-x}$. Same result.

   ✓Final answer

   $\dfrac{dy}{dx} = -e^{-x}$, i.e. option (A).

   ANSWER: A

5. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.If $f(x) = \dfrac{e^{-x} \sin x}{\log_e x}$ and $f'(x) = f(x) \cdot g(x)$, then $g'(e) =$ (A) $e^{-2} - \csc^2(e)$ (B) $2e^{2} - \csc^2(e)$ (C) $2e^{-2} - \csc^2(e)$ (D) $2e^{-2} + \csc^2(e)$

   ›Reveal solutionSolution

   We use logarithmic differentiation to simplify $f(x)$ into a sum of terms, which directly gives us $g(x) = \frac{f'(x)}{f(x)}$. Differentiating $g(x)$ and substituting $x=e$ then yields the result. The value of $g'(e)$ is $\boxed{2e^{-2} - \csc^2(e)}$.

   The problem asks us to find $g'(e)$ given a function $f(x)$ and the relationship $f'(x) = f(x) \cdot g(x)$. The function $f(x)$ is a product and quotient of several functions, making direct differentiation quite cumbersome.

   The key insight here is to recognize that the expression $g(x) = \frac{f'(x)}{f(x)}$ is precisely the derivative of $\log_e |f(x)|$. This means we can use logarithmic differentiation to find $g(x)$ efficiently. By taking the natural logarithm of $f(x)$ first, we convert products and quotients into sums and differences, which are much simpler to differentiate.

   Here's how we approach the problem:

   1. Express $g(x)$ using logarithmic differentiation:

      Given $f'(x) = f(x) \cdot g(x)$, we can write $g(x) = \frac{f'(x)}{f(x)}$.

      This expression is the result of differentiating $\log_e f(x)$ with respect to $x$.

      So, our first step is to take the natural logarithm of $f(x)$ and then differentiate it.

      We have $f(x) = \dfrac{e^{-x} \sin x}{\log_e x}$.

      Taking the natural logarithm on both sides:

   $\log_e f(x) = \log_e \left( \dfrac{e^{-x} \sin x}{\log_e x} \right)$

   Using the properties of logarithms ($\log(AB/C) = \log A + \log B - \log C$):
   

   $\log_e f(x) = \log_e (e^{-x}) + \log_e (\sin x) - \log_e (\log_e x)$

   Simplify the first term: $\log_e (e^{-x}) = -x$.
   

   $\log_e f(x) = -x + \log_e (\sin x) - \log_e (\log_e x)$

   2. Differentiate to find $g(x)$: Now, differentiate both sides of the equation with respect to $x$:

   $\frac{d}{dx} (\log_e f(x)) = \frac{d}{dx} (-x) + \frac{d}{dx} (\log_e (\sin x)) - \frac{d}{dx} (\log_e (\log_e x))$

   We know that $\frac{d}{dx} (\log_e f(x)) = \frac{f'(x)}{f(x)}$, which is $g(x)$.
   Differentiating each term on the right side:
   *   $\frac{d}{dx} (-x) = -1$
   *   $\frac{d}{dx} (\log_e (\sin x)) = \frac{1}{\sin x} \cdot \cos x = \cot x$
   *   $\frac{d}{dx} (\log_e (\log_e x)) = \frac{1}{\log_e x} \cdot \frac{1}{x}$ (using the chain rule)
   
   Combining these, we get $g(x)$:
   

   $g(x) = -1 + \cot x - \frac{1}{x \log_e x}$

   3. Differentiate $g(x)$ to find $g'(x)$: Now we need to find the derivative of $g(x)$:

   $g'(x) = \frac{d}{dx} (-1) + \frac{d}{dx} (\cot x) - \frac{d}{dx} \left( \frac{1}{x \log_e x} \right)$

   *   $\frac{d}{dx} (-1) = 0$
   *   $\frac{d}{dx} (\cot x) = -\csc^2 x$
   
   For the last term, $\frac{d}{dx} \left( \frac{1}{x \log_e x} \right)$, we can use the quotient rule or chain rule. Let $u = x \log_e x$. Then we are differentiating $u^{-1}$.
   $\frac{d}{dx} (u^{-1}) = -1 \cdot u^{-2} \cdot \frac{du}{dx} = - \frac{1}{u^2} \cdot \frac{du}{dx}$.
   First, find $\frac{du}{dx} = \frac{d}{dx} (x \log_e x)$ using the product rule:
   

   $\frac{d}{dx} (x \log_e x) = (1 \cdot \log_e x) + (x \cdot \frac{1}{x}) = \log_e x + 1$

   So,
   

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

   Substituting these back into the expression for $g'(x)$:
   

   $g'(x) = 0 - \csc^2 x - \left( - \frac{\log_e x + 1}{(x \log_e x)^2} \right)$

   $g'(x) = -\csc^2 x + \frac{\log_e x + 1}{(x \log_e x)^2}$

   4. Substitute $x=e$ into $g'(x)$: Finally, substitute $x=e$ into the expression for $g'(x)$. Recall that $\log_e e = 1$.

   $g'(e) = -\csc^2 e + \frac{\log_e e + 1}{(e \log_e e)^2}$

   $g'(e) = -\csc^2 e + \frac{1 + 1}{(e \cdot 1)^2}$

   $g'(e) = -\csc^2 e + \frac{2}{e^2}$

   This can be written as:
   

   $g'(e) = 2e^{-2} - \csc^2 e$

   Comparing this result with the given options, it matches option (C).

   ✓Final answer

   The value of $g'(e)$ is $\boxed{2e^{-2} - \csc^2(e)}$.

6. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.If $f(x) = \dfrac{\cos^2 x}{1 + \sin^2 x}$, then $f\left(\dfrac{\pi}{4}\right) - 3 f'\left(\dfrac{\pi}{4}\right) =$ (A) $\dfrac{5}{3}$ (B) $\dfrac{11}{3}$ (C) $\dfrac{13}{9}$ (D) $3$

   ›Reveal solutionSolution

   $f\!\left(\tfrac\pi4\right)=\tfrac13$ and $f'\!\left(\tfrac\pi4\right)=-\tfrac89$, so $f\!\left(\tfrac\pi4\right)-3f'\!\left(\tfrac\pi4\right)=\tfrac13+\tfrac{24}{9}=3$ — option (D).

   Evaluate $f(\pi/4)$. With $f(x)=\dfrac{\cos^2x}{1+\sin^2x}$ and $\cos^2\tfrac\pi4=\sin^2\tfrac\pi4=\tfrac12$:

   $f\!\left(\tfrac\pi4\right)=\frac{1/2}{1+1/2}=\frac{1/2}{3/2}=\frac13.$

   Differentiate. With $u=\cos^2x,\;v=1+\sin^2x$ (so $u'=-\sin2x,\;v'=\sin2x$):

   $f'(x)=\frac{u'v-uv'}{v^2}=\frac{-\sin2x\,(1+\sin^2x)-\cos^2x\,\sin2x}{(1+\sin^2x)^2}=\frac{-\sin2x\,(2)}{(1+\sin^2x)^2}=\frac{-2\sin2x}{(1+\sin^2x)^2}.$

   At $x=\tfrac\pi4$: $\sin2x=1$ and $1+\sin^2\tfrac\pi4=\tfrac32$, so

   $f'\!\left(\tfrac\pi4\right)=\frac{-2}{(3/2)^2}=\frac{-2}{9/4}=-\frac89.$

   Combine.

   $f\!\left(\tfrac\pi4\right)-3f'\!\left(\tfrac\pi4\right)=\frac13-3\!\left(-\frac89\right)=\frac13+\frac{24}{9}=\frac13+\frac83=3.$

   ✓Final answer

   $f\!\left(\dfrac\pi4\right)-3f'\!\left(\dfrac\pi4\right)=3$ — option (D).

7. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $f(x) = (1 + x^3)(1 + x^6)(1 + x^{12})(1 + x^{24})$, then $f'(-1) =$ (A) 24 (B) 12 (C) 48 (D) 60

   ›Reveal solutionSolution

   At $x=-1$ the factor $(1+x^3)$ vanishes, so only the term where it is differentiated survives: $f'(-1)=24$ — option (A).

   For a product $f=f_1 f_2 f_3 f_4$, the derivative is $f'=f_1'f_2f_3f_4+f_1f_2'f_3f_4+\cdots$. Every term keeps three of the original factors undifferentiated.

   1. Note the vanishing factor. At $x=-1$, $\;1+x^3=1+(-1)^3=0$. Any product-rule term that still contains the factor $(1+x^3)$ is therefore $0$. Only the single term in which $(1+x^3)$ is the one being differentiated can be non-zero.

   2. Keep the surviving term.

   $f'(x)\Big|_{\text{surviving}}=\frac{d}{dx}(1+x^3)\cdot(1+x^6)(1+x^{12})(1+x^{24})=3x^2\,(1+x^6)(1+x^{12})(1+x^{24}).$

   3. Evaluate at $x=-1$. Here $(-1)^6=(-1)^{12}=(-1)^{24}=1$, so each remaining factor equals $2$, and $3x^2=3(1)=3$:

   $f'(-1)=3\cdot(2)(2)(2)=3\cdot 8=24.$

   ✓Final answer

   $f'(-1)=24$ — option (A).

8. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.Match the values of $\dfrac{dy}{dx}$ at $x = \dfrac{\pi}{3}$ for the following system of curves in parametric form given in List-I with those of the items in List-II List-I

   i) $x = a(\theta - \sin \theta),\ y = a(1 - \cos \theta)$

   ii) $x = 3\cos \theta - 2\cos^3 \theta,\ y = 3\sin \theta - 2\sin^3 \theta$

   iii) $x = 3\cos \theta - \cos^3 \theta,\ y = 3\sin \theta - \sin^3 \theta$

   iv) $x = a \log \sin \theta,\ y = a \tan \theta$ List-II

   a) $4\sqrt{3}$

   b) $-\dfrac{1}{3\sqrt{3}}$

   c) $\sqrt{3}$

   d) $\dfrac{1}{\sqrt{3}}$

   e) $\dfrac{1}{3\sqrt{3}}$ (A)

   (i) $\rightarrow$ c,

   (ii) $\rightarrow$ d,

   (iii) $\rightarrow$ b,

   (iv) $\rightarrow$ a (B)

   (i) $\rightarrow$ c,

   (ii) $\rightarrow$ e,

   (iii) $\rightarrow$ d,

   (iv) $\rightarrow$ a (C)

   (i) $\rightarrow$ d,

   (ii) $\rightarrow$ c,

   (iii) $\rightarrow$ b,

   (iv) $\rightarrow$ a (D)

   (i) $\rightarrow$ d,

   (ii) $\rightarrow$ c,

   (iii) $\rightarrow$ e,

   (iv) $\rightarrow$ b

   ›Reveal solutionSolution

   Each slope simplifies to a clean trig expression — $\cot(\theta/2)$, $\cot\theta$, $-\cot^{3}\theta$ and $\sec^{2}\theta\tan\theta$ — evaluated at $\theta=\pi/3$ they give $\sqrt3,\ \tfrac{1}{\sqrt3},\ -\tfrac{1}{3\sqrt3},\ 4\sqrt3$. That is i→c, ii→d, iii→b, iv→a — option (A).

   The concept first: simplify symbolically, substitute last

   When $x$ and $y$ are both functions of a parameter $\theta$, the chain rule gives

   $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\qquad (\text{provided } dx/d\theta \neq 0).$

   The temptation is to plug $\theta = \pi/3$ into $dy/d\theta$ and $dx/d\theta$ immediately. Resist it: in every one of these four curves an enormous common factor cancels (a $\cos 2\theta$, a $3\sin\theta$, an $a$…), and the ratio collapses to something you can evaluate in your head. Simplify first, substitute last — that is the entire craft here.

   (i) Cycloid: $x = a(\theta - \sin\theta),\ y = a(1-\cos\theta)$

   $\frac{dx}{d\theta} = a(1 - \cos\theta), \qquad \frac{dy}{d\theta} = a\sin\theta$

   $\frac{dy}{dx} = \frac{a\sin\theta}{a(1-\cos\theta)} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^{2}\frac{\theta}{2}} = \cot\frac{\theta}{2}$

   At $\theta = \pi/3$: $\cot\dfrac{\pi}{6} = \sqrt3$. ⇒ (i) → c

   (ii) $x = 3\cos\theta - 2\cos^{3}\theta,\ y = 3\sin\theta - 2\sin^{3}\theta$

   $\frac{dx}{d\theta} = -3\sin\theta + 6\cos^{2}\theta\sin\theta = 3\sin\theta\,(2\cos^{2}\theta - 1) = 3\sin\theta\cos 2\theta$

   $\frac{dy}{d\theta} = 3\cos\theta - 6\sin^{2}\theta\cos\theta = 3\cos\theta\,(1 - 2\sin^{2}\theta) = 3\cos\theta\cos 2\theta$

   The $\cos 2\theta$ cancels beautifully:

   $\frac{dy}{dx} = \cot\theta \quad\Rightarrow\quad \cot\frac{\pi}{3} = \frac{1}{\sqrt3}$

   ⇒ (ii) → d

   (iii) $x = 3\cos\theta - \cos^{3}\theta,\ y = 3\sin\theta - \sin^{3}\theta$

   $\frac{dx}{d\theta} = -3\sin\theta + 3\cos^{2}\theta\sin\theta = -3\sin\theta(1 - \cos^{2}\theta) = -3\sin^{3}\theta$

   $\frac{dy}{d\theta} = 3\cos\theta - 3\sin^{2}\theta\cos\theta = 3\cos\theta(1 - \sin^{2}\theta) = 3\cos^{3}\theta$

   $\frac{dy}{dx} = \frac{3\cos^{3}\theta}{-3\sin^{3}\theta} = -\cot^{3}\theta$

   At $\theta = \pi/3$: $\cot\dfrac{\pi}{3} = \dfrac{1}{\sqrt3}$, so $-\left(\dfrac{1}{\sqrt3}\right)^{3} = -\dfrac{1}{3\sqrt3}$. ⇒ (iii) → b

   (Note how the single coefficient in (iii) versus the 2 in (ii) completely changes the answer — that contrast is the point of pairing these two curves.)

   (iv) $x = a\log\sin\theta,\ y = a\tan\theta$

   $\frac{dx}{d\theta} = a\cdot\frac{\cos\theta}{\sin\theta} = a\cot\theta, \qquad \frac{dy}{d\theta} = a\sec^{2}\theta$

   $\frac{dy}{dx} = \frac{a\sec^{2}\theta}{a\cot\theta} = \sec^{2}\theta\,\tan\theta$

   At $\theta = \pi/3$: $\sec\dfrac{\pi}{3} = 2 \Rightarrow \sec^{2} = 4$, and $\tan\dfrac{\pi}{3} = \sqrt3$:

   $\frac{dy}{dx} = 4\sqrt3$

   ⇒ (iv) → a

   Collect: i→c, ii→d, iii→b, iv→a.

   ✓Final answer

   The correct matching is i→c, ii→d, iii→b, iv→a, so the correct option is (A).

   ANSWER: A

9. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.The derivate of $(\log x)^{\sin x}$ with respect to $\cos x$ at $x = \dfrac{\pi}{2}$ is (A) $\dfrac{-4}{\pi}$ (B) $\dfrac{-\pi}{2}$ (C) $\dfrac{-2}{\pi}$ (D) $\dfrac{-\pi}{4}$

   ›Reveal solutionSolution

   To find the derivative of $u$ with respect to $v$ when both are functions of $x$, we use the chain rule: $\frac{du}{dv} = \frac{du/dx}{dv/dx}$. We apply logarithmic differentiation to find $\frac{du}{dx}$ for $u = (\log x)^{\sin x}$, and then evaluate the expression at $x = \frac{\pi}{2}$. The result is $\boxed{\dfrac{-2}{\pi}}$.

   When asked to find the derivative of one function, say $u$, with respect to another function, say $v$, and both $u$ and $v$ are themselves functions of a third variable, say $x$, we use a specific application of the chain rule. This is often called parametric differentiation.

   The core idea is that if $u = f(x)$ and $v = g(x)$, then the derivative of $u$ with respect to $v$ is given by:

   $\frac{du}{dv} = \frac{du/dx}{dv/dx}$

   provided $\frac{dv}{dx} \neq 0$.

   In this problem, we have $u = (\log x)^{\sin x}$ and $v = \cos x$. We need to find $\frac{du}{dv}$ at $x = \frac{\pi}{2}$.

   Here's how we approach it:

   1. Define the functions:

      Let $u = (\log x)^{\sin x}$ and $v = \cos x$.

      Our goal is to find $\frac{du}{dv}$ at $x = \frac{\pi}{2}$.

   2. Find $\frac{du}{dx}$ using logarithmic differentiation:

      The function $u = (\log x)^{\sin x}$ is of the form $f(x)^{g(x)}$, which is best differentiated using logarithms.

      Take the natural logarithm on both sides:

   $\log u = \log \left( (\log x)^{\sin x} \right)$

   Using the logarithm property $\log(a^b) = b \log a$:
   

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

   Now, differentiate both sides with respect to $x$. Remember to use the product rule on the right side and the chain rule on the left side.
   

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

   We know $\frac{d}{dx}(\sin x) = \cos x$.
   For $\frac{d}{dx}(\log(\log x))$, we apply the chain rule: $\frac{d}{dx}(\log(f(x))) = \frac{1}{f(x)} f'(x)$. Here, $f(x) = \log x$, so $f'(x) = \frac{1}{x}$.
   

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

   Substitute these derivatives back into the equation:
   

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

   Now, solve for $\frac{du}{dx}$:
   

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

   Substitute $u = (\log x)^{\sin x}$ back:
   

   $\frac{du}{dx} = (\log x)^{\sin x} \left( \cos x \log(\log x) + \frac{\sin x}{x \log x} \right)$

   3. Find $\frac{dv}{dx}$: The function $v = \cos x$ is straightforward to differentiate:

   $\frac{dv}{dx} = -\sin x$

   4. Apply the chain rule $\frac{du}{dv} = \frac{du/dx}{dv/dx}$:

   $\frac{du}{dv} = \frac{(\log x)^{\sin x} \left( \cos x \log(\log x) + \frac{\sin x}{x \log x} \right)}{-\sin x}$

   5. Evaluate at $x = \frac{\pi}{2}$:

      Now, substitute $x = \frac{\pi}{2}$ into the expression for $\frac{du}{dv}$.

      Recall the values of trigonometric functions at $x = \frac{\pi}{2}$:

      $\sin\left(\frac{\pi}{2}\right) = 1$

      $\cos\left(\frac{\pi}{2}\right) = 0$

      Let's evaluate the numerator first:

   $\left. (\log x)^{\sin x} \left( \cos x \log(\log x) + \frac{\sin x}{x \log x} \right) \right|_{x=\frac{\pi}{2}}$

   $= \left( \log\left(\frac{\pi}{2}\right) \right)^{\sin(\frac{\pi}{2})} \left( \cos\left(\frac{\pi}{2}\right) \log\left(\log\left(\frac{\pi}{2}\right)\right) + \frac{\sin\left(\frac{\pi}{2}\right)}{\frac{\pi}{2} \log\left(\frac{\pi}{2}\right)} \right)$

   $= \left( \log\left(\frac{\pi}{2}\right) \right)^1 \left( 0 \cdot \log\left(\log\left(\frac{\pi}{2}\right)\right) + \frac{1}{\frac{\pi}{2} \log\left(\frac{\pi}{2}\right)} \right)$

   $= \log\left(\frac{\pi}{2}\right) \left( 0 + \frac{1}{\frac{\pi}{2} \log\left(\frac{\pi}{2}\right)} \right)$

   $= \log\left(\frac{\pi}{2}\right) \cdot \frac{1}{\frac{\pi}{2} \log\left(\frac{\pi}{2}\right)}$

   The $\log\left(\frac{\pi}{2}\right)$ terms cancel out (since $\frac{\pi}{2} \approx 1.57$, $\log(\frac{\pi}{2})$ is a non-zero positive value).
   

   $= \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi}$

   Now, evaluate the denominator at $x = \frac{\pi}{2}$:
   

   $\left. -\sin x \right|_{x=\frac{\pi}{2}} = -\sin\left(\frac{\pi}{2}\right) = -1$

   Finally, combine the numerator and denominator:
   

   $\left. \frac{du}{dv} \right|_{x=\frac{\pi}{2}} = \frac{\frac{2}{\pi}}{-1} = -\frac{2}{\pi}$

   The correct option is (C).

   ✓Final answer

   The derivative of $(\log x)^{\sin x}$ with respect to $\cos x$ at $x = \frac{\pi}{2}$ is $\boxed{\dfrac{-2}{\pi}}$.

10. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.If $y = \cos^{-1}(\tanh x) + \sinh(\sin 6x)$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{-1}{\cosh x} + 6 \cos 6x \cosh(\sin 6x)$ (B) $\dfrac{1}{\cosh x} - 6 \cos 6x \cosh(\sin 6x)$ (C) $\dfrac{-1}{\cosh x} - 6 \cos 6x \cosh(\sin 6x)$ (D) $\dfrac{1}{\cosh x} + 6 \cos 6x \cosh(\sin 6x)$

    ›Reveal solutionSolution

    Differentiate each term separately using chain rule and known derivatives: derivative of $\cos^{-1}(\tanh x)$ is $-\frac{1}{\cosh x}$, and derivative of $\sinh(\sin 6x)$ is $6\cos 6x \cosh(\sin 6x)$. The sum gives option (A).

    The function is a sum of two completely different pieces: an inverse cosine of a hyperbolic tangent, and a hyperbolic sine of a sine. Each requires its own chain rule application, and the derivatives never mix. The key is to handle them one at a time, keeping the algebra clean.

    1. First term: $y_1 = \cos^{-1}(\tanh x)$ Recall: $\frac{d}{du} \cos^{-1} u = \frac{-1}{\sqrt{1-u^2}}$. Here $u = \tanh x$, so by the chain rule:

    $\frac{dy_1}{dx} = \frac{-1}{\sqrt{1 - \tanh^2 x}} \cdot \frac{d}{dx}(\tanh x).$

    Now $\frac{d}{dx}(\tanh x) = \text{sech}^2 x = \frac{1}{\cosh^2 x}$.

    Also, $1 - \tanh^2 x = \text{sech}^2 x = \frac{1}{\cosh^2 x}$, so $\sqrt{1 - \tanh^2 x} = \frac{1}{\cosh x}$ (taking the positive root since $\cosh x > 0$).

    Therefore:

    $\frac{dy_1}{dx} = \frac{-1}{1/\cosh x} \cdot \frac{1}{\cosh^2 x} = -\cosh x \cdot \frac{1}{\cosh^2 x} = -\frac{1}{\cosh x}.$

    2. Second term: $y_2 = \sinh(\sin 6x)$ Recall: $\frac{d}{du} \sinh u = \cosh u$. With $u = \sin 6x$, chain rule gives:

    $\frac{dy_2}{dx} = \cosh(\sin 6x) \cdot \frac{d}{dx}(\sin 6x).$

    And $\frac{d}{dx}(\sin 6x) = 6 \cos 6x$.

    So:

    $\frac{dy_2}{dx} = 6 \cos 6x \cdot \cosh(\sin 6x).$

    3. Combine:

    $\frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} = -\frac{1}{\cosh x} + 6 \cos 6x \cosh(\sin 6x).$

    Watch out

    A common slip is forgetting the minus sign from the derivative of $\cos^{-1}$, or mixing up $\cosh$ and $\text{sech}$ in the simplification. Always write $1 - \tanh^2 x = \text{sech}^2 x$ explicitly to avoid errors.

    ✓Final answer

    The correct option is (A).

11. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.If $y = \tan^{-1} \left[ \left( \frac{1 - \cos 2\sqrt{x}}{1 + \cos 2\sqrt{x}} \right)^{1/2} \right],\ 0 < x < \frac{\pi^2}{4}$, then $y(2y' + y) =$ (A) $1$ (B) $x + 1$ (C) $\sqrt{x}$ (D) $\sqrt{x} + 1$

    ›Reveal solutionSolution

    Simplify the argument with half-angle identities to get $y=\sqrt{x}$, so $y'=\dfrac{1}{2\sqrt{x}}$ and $y(2y'+y)=x+1$ — option (B).

    Simplify the inside first. Using $1-\cos 2\theta = 2\sin^2\theta$ and $1+\cos 2\theta = 2\cos^2\theta$ with $\theta=\sqrt{x}$:

    $\frac{1-\cos 2\sqrt{x}}{1+\cos 2\sqrt{x}} = \frac{2\sin^2\sqrt{x}}{2\cos^2\sqrt{x}} = \tan^2\sqrt{x}.$

    Taking the square root gives $\left(\tan^2\sqrt{x}\right)^{1/2}=|\tan\sqrt{x}|$. For $0<x<\tfrac{\pi^2}{4}$ we have $0<\sqrt{x}<\tfrac{\pi}{2}$, so $\tan\sqrt{x}>0$ and

    $y=\tan^{-1}\!\left(\tan\sqrt{x}\right)=\sqrt{x},$

    since $\sqrt{x}$ lies in the principal range $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ of $\tan^{-1}$.

    Differentiate. With $y=x^{1/2}$,

    $y'=\frac{1}{2\sqrt{x}}.$

    Evaluate the required expression.

    $2y'+y = 2\cdot\frac{1}{2\sqrt{x}}+\sqrt{x} = \frac{1}{\sqrt{x}}+\sqrt{x},$

    $y(2y'+y) = \sqrt{x}\left(\frac{1}{\sqrt{x}}+\sqrt{x}\right) = 1 + x.$

    ✓Final answer

    $y(2y'+y) = x+1$ — option (B).

12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.If $x = \sqrt{1 - \tan y}$, then $\dfrac{dy}{dx} =$ (A) $-\dfrac{2x}{x^4 + 2x^2 + 2}$ (B) $-\dfrac{2x}{x^4 - 2x^2 + 2}$ (C) $\dfrac{2x}{x^4 - 2x^2 + 2}$ (D) $\dfrac{2x}{x^4 + 2x^2 + 2}$

    ›Reveal solutionSolution

    Implicit differentiation gives $\dfrac{dy}{dx}=-\dfrac{2x}{x^{4}-2x^{2}+2}$. Option (B).

    Solution

    From $x=\sqrt{1-\tan y}$, square both sides:

    $x^{2}=1-\tan y\ \Longrightarrow\ \tan y=1-x^{2}.$

    Differentiate $x^{2}=1-\tan y$ with respect to $x$:

    $2x=-\sec^{2}y\,\frac{dy}{dx}\ \Longrightarrow\ \frac{dy}{dx}=-\frac{2x}{\sec^{2}y}.$

    Express $\sec^{2}y$ in terms of $x$ using $\sec^{2}y=1+\tan^{2}y$:

    $\sec^{2}y=1+(1-x^{2})^{2}=1+1-2x^{2}+x^{4}=x^{4}-2x^{2}+2.$

    Therefore

    $\frac{dy}{dx}=-\frac{2x}{x^{4}-2x^{2}+2}.$

    ✓Final answer

    Option (B): $\;-\dfrac{2x}{x^{4}-2x^{2}+2}$.