Q.Differentiate w.r.t. $x$, the following function:

Showing the 12 most recent of 26 on this concept.

1. 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 $\boxed{\dfrac{-2}{\pi}}$.

2. TG EAPCET 2025Set eng-2025-05-03-AN1 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

   Square the given relation to get $\tan y=1-x^{2}$, i.e. $y=\tan^{-1}(1-x^{2})$, and differentiate: $\dfrac{dy}{dx}=-\dfrac{2x}{x^{4}-2x^{2}+2}$ — option (B).

   The concept first

   When $y$ is buried inside a trigonometric function and $x$ sits outside a radical, do not rush into implicit differentiation. It is far cleaner to tidy the relation algebraically first, so that $y$ is written explicitly in terms of $x$; then a single application of the chain rule finishes it. The tool you need is

   $\frac{d}{dx}\tan^{-1}u=\frac{1}{1+u^{2}}\cdot\frac{du}{dx}.$

   Step 1 — Make $y$ explicit

   $x=\sqrt{1-\tan y}\;\Longrightarrow\;x^{2}=1-\tan y\;\Longrightarrow\;\tan y=1-x^{2}\;\Longrightarrow\;y=\tan^{-1}\!\left(1-x^{2}\right).$

   Step 2 — Differentiate with the chain rule

   With $u=1-x^{2}$, $\dfrac{du}{dx}=-2x$:

   $\frac{dy}{dx}=\frac{1}{1+\left(1-x^{2}\right)^{2}}\times(-2x)=\frac{-2x}{1+\left(1-x^{2}\right)^{2}} .$

   Step 3 — Simplify the denominator

   $\left(1-x^{2}\right)^{2}=1-2x^{2}+x^{4},$

   $1+\left(1-x^{2}\right)^{2}=1+1-2x^{2}+x^{4}=x^{4}-2x^{2}+2 .$

   Therefore

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

   Step 4 — Check the sign makes sense

   As $x$ increases (for $x>0$), $\tan y=1-x^{2}$ decreases, so $y$ must decrease — the derivative should be negative. Our answer is negative for $x>0$ (the denominator $x^{4}-2x^{2}+2=(x^{2}-1)^{2}+1>0$ always). ✓ This also rules out option (C) immediately; options (A) and (D) carry the wrong denominator, $x^4+2x^2+2$.

   ✓Final answer

   $\dfrac{dy}{dx}=-\dfrac{2x}{x^{4}-2x^{2}+2}$, which is option (B).

   ANSWER: B

3. 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): $\;-\dfrac{2x}{x^{4}-2x^{2}+2}$.

4. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $\sec \left( \log_2 y^2 \right) = \csc \left( \log_2 x^2 \right)$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{x}{y}$ (B) $\dfrac{y}{x}$ (C) $-\dfrac{y}{x}$ (D) $-\dfrac{x}{y}$

   ›Reveal solutionSolution

   The key idea is to rewrite the given equation using the identity $\sec\theta = \csc(\frac{\pi}{2} - \theta)$, then equate the arguments of the logs (up to an additive constant) and differentiate implicitly. The result is $\frac{dy}{dx} = -\frac{y}{x}$, which corresponds to option (C).

   ---

   We start with

   $\sec\left(\log_2 y^2\right) = \csc\left(\log_2 x^2\right).$

   Concept and intuition

   The equation mixes secant and cosecant of different arguments. A natural way to compare them is to use the cofunction identity:

   $\sec A = \csc\left(\frac{\pi}{2} - A\right).$

   This lets us rewrite the left-hand side as a cosecant, so both sides become cosecants of some expressions. Then, because cosecant is not one-to-one over all reals, we must consider that equality of cosecants means their arguments differ by an integer multiple of $2\pi$ or are supplementary (since $\csc \alpha = \csc \beta$ implies $\alpha = \beta + 2\pi n$ or $\alpha = \pi - \beta + 2\pi n$). However, the presence of logs and the fact that $x$ and $y$ are variables (likely positive, so logs are defined) suggests the simplest branch will give the relation we need. We’ll assume the principal branch and later check that the derivative is independent of the integer constant.

   ---

   Step-by-step solution

   1. Apply the cofunction identity

   $\sec\left(\log_2 y^2\right) = \csc\left(\frac{\pi}{2} - \log_2 y^2\right).$

   So the equation becomes

   $\csc\left(\frac{\pi}{2} - \log_2 y^2\right) = \csc\left(\log_2 x^2\right).$

   2. Equate the arguments (up to periodicity) For cosecant, $\csc \alpha = \csc \beta$ implies

   $\alpha = \beta + 2\pi k \quad \text{or} \quad \alpha = \pi - \beta + 2\pi k,$

   for some integer $k$.

   The second case would introduce a constant shift that, upon differentiation, disappears anyway. So we take the simplest:

   $\frac{\pi}{2} - \log_2 y^2 = \log_2 x^2 + C,$

   where $C$ is a constant (combining the $2\pi k$ or $\pi$ shift).

   For differentiation, any constant $C$ will vanish.

   3. Simplify the logs Recall $\log_2 y^2 = 2\log_2 y$ and $\log_2 x^2 = 2\log_2 x$. So

   $\frac{\pi}{2} - 2\log_2 y = 2\log_2 x + C.$

   4. Differentiate implicitly with respect to $x$ Differentiate term by term:

   $0 - 2\cdot\frac{1}{y\ln 2}\cdot\frac{dy}{dx} = 2\cdot\frac{1}{x\ln 2} + 0.$

   (The derivative of the constant $C$ is $0$, and $\frac{d}{dx}\log_2 y = \frac{1}{y\ln 2}\frac{dy}{dx}$.)

   5. Solve for $\frac{dy}{dx}$ Multiply both sides by $\ln 2$:

   $-\frac{2}{y}\frac{dy}{dx} = \frac{2}{x}.$

   Divide by 2:

   $-\frac{1}{y}\frac{dy}{dx} = \frac{1}{x}.$

   Multiply both sides by $-y$:

   $\frac{dy}{dx} = -\frac{y}{x}.$

   ---

   Tip

   The constant $C$ (from the periodic nature of cosecant) disappears upon differentiation, so the result is robust regardless of which branch we pick. This is a common trick: implicit differentiation of trigonometric equations often yields a derivative independent of additive constants.

   Watch out

   A classic mistake is to forget the factor of 2 from $\log_2 y^2 = 2\log_2 y$, leading to an incorrect factor. Always simplify logs before differentiating.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

5. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.If $y = \cos^{-1} \left( \dfrac{6x - 2x^2 - 4}{2x^2 - 6x + 5} \right)$ then $\dfrac{dy}{dx} =$ (A) $\dfrac{2}{\sqrt{3x - x^2 - 2}}$ (B) $\dfrac{2}{3x - x^2 - 2}$ (C) $\dfrac{2}{\sqrt{2x^2 - 6x + 5}}$ (D) $\dfrac{2}{2x^2 - 6x + 5}$

   ›Reveal solutionSolution

   With $t=2x^2-6x+4$ the argument is $\dfrac{-t}{t+1}$, and the derivative collapses to $\dfrac{2}{2x^2-6x+5}$.

   Write $y=\cos^{-1}u$ where

   $u=\frac{6x-2x^2-4}{2x^2-6x+5}=\frac{-(2x^2-6x+4)}{(2x^2-6x+4)+1}.$

   Let $t=2x^2-6x+4$, so $u=\dfrac{-t}{t+1}$ and $t+1=2x^2-6x+5$.

   Compute $1-u^2$.

   $1-u^2=\frac{(t+1)^2-t^2}{(t+1)^2}=\frac{2t+1}{(t+1)^2}.$

   Now $2t+1=2(2x^2-6x+4)+1=4x^2-12x+9=(2x-3)^2$, hence

   $\sqrt{1-u^2}=\frac{|2x-3|}{\,t+1\,}\qquad(t+1>0\text{ always, since its discriminant }36-40<0).$

   Differentiate $u$. Since $t'=4x-6=2(2x-3)$,

   $\frac{du}{dx}=\frac{-t'(t+1)+t\,t'}{(t+1)^2}=\frac{-t'}{(t+1)^2}=\frac{-2(2x-3)}{(t+1)^2}.$

   Chain rule.

   $\frac{dy}{dx}=-\frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx}=-\frac{t+1}{|2x-3|}\cdot\frac{-2(2x-3)}{(t+1)^2}=\frac{2}{\,t+1\,}=\frac{2}{2x^2-6x+5}.$

   ✓Final answer

   $\dfrac{dy}{dx}=\dfrac{2}{2x^2-6x+5}$ — option (D).

6. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.If $y = \sqrt{\sin(\log 2x)} + \sqrt{\sin(\log 2x)} + \sqrt{\sin(\log 2x)} + \ldots \infty$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{\cos(\log 2x)}{2x(2y-1)}$ (B) $\dfrac{\cos(\log 2x)}{(2y-1)}$ (C) $\dfrac{\cos(\log 2x)}{x(2y-1)}$ (D) $\dfrac{\sin(\log 2x)}{x(2y-1)}$

   ›Reveal solutionSolution

   The infinite sum collapses to a simple equation $y = \sqrt{\sin(\log 2x)} + y$, which forces us to reinterpret the expression as a self-repeating pattern. The correct interpretation is $y = \sqrt{\sin(\log 2x) + \sqrt{\sin(\log 2x) + \dots}}$, leading to $y^2 = \sin(\log 2x) + y$, and differentiating gives $\frac{dy}{dx} = \frac{\cos(\log 2x)}{x(2y-1)}$, so the answer is (C).

   ---

   The key here is to first understand what the infinite expression actually means. At first glance, it looks like a sum of identical terms: $\sqrt{\sin(\log 2x)} + \sqrt{\sin(\log 2x)} + \dots$ to infinity. But that sum would diverge (unless the term is zero), so it cannot be that. Instead, the notation is a classic trick: it means an infinite nested radical, where each radical contains the entire rest of the expression. That is:

   $y = \sqrt{\sin(\log 2x) + \sqrt{\sin(\log 2x) + \sqrt{\sin(\log 2x) + \dots}}}$

   This is a self-similar structure: the whole expression appears again inside itself. That self-reference lets us write a simple algebraic equation for $y$.

   ---

   1. Write the self-referential equation Since the expression inside the first square root is exactly the same as the whole $y$, we have:

   $y = \sqrt{\sin(\log 2x) + y}$

   This is the crucial step — it turns an infinite process into a finite equation.

   2. Square both sides

   $y^2 = \sin(\log 2x) + y$

   Rearranging:

   $y^2 - y = \sin(\log 2x)$

   3. Differentiate implicitly with respect to $x$ Differentiate both sides:

   $2y \frac{dy}{dx} - \frac{dy}{dx} = \cos(\log 2x) \cdot \frac{1}{2x} \cdot 2$

   The derivative of $\sin(\log 2x)$ uses the chain rule: derivative of $\sin$ is $\cos$, derivative of $\log 2x$ is $\frac{1}{2x} \cdot 2 = \frac{1}{x}$. So:

   $(2y - 1) \frac{dy}{dx} = \frac{\cos(\log 2x)}{x}$

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

   $\frac{dy}{dx} = \frac{\cos(\log 2x)}{x(2y - 1)}$

   This matches option (C).

   Watch out

   A common mistake is to treat the given expression as an infinite sum of identical radicals. That sum would be infinite unless the term is zero, which is not the case here. Always check: if it were a sum, the problem would be meaningless. The notation $\sqrt{\dots} + \sqrt{\dots} + \dots$ is a misdirection — the intended meaning is a nested radical.

   Tip

   The self-referential trick $y = \sqrt{\text{something} + y}$ is a standard method for infinite nested radicals. It works because the pattern repeats identically at every level.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

7. 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 $\boxed{2e^{-2} - \csc^2(e)}$.

8. 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).

9. 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

10. 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).

11. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.$\lim_{x \to 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{1 - \cos 4x} =$ (A) $\frac{3}{5} (\log 18 - 1)$ (B) $\frac{5}{16} \log \left(\frac{81}{e}\right)$ (C) $\frac{4}{15} (\log 81 - 1)$ (D) $\frac{16}{5} [\log(27) - 1]$

    ›Reveal solutionSolution

    The limit is evaluated by expanding each term near $x=0$ using series expansions (or standard limits) and simplifying; the result matches option (B).

    We need to compute

    $L = \lim_{x \to 0} \frac{(3^{2x} - \sqrt{x+1}) \sin 5x}{1 - \cos 4x}.$

    The denominator and numerator both vanish at $x=0$, so this is a $\frac{0}{0}$ form. The key is to replace each piece with its leading-order behaviour near $x=0$.

    ---

    1. Expand the denominator

    Recall the standard limit

    $1 - \cos u \sim \frac{u^2}{2} \quad \text{as } u \to 0.$

    Here $u = 4x$, so

    $1 - \cos 4x \sim \frac{(4x)^2}{2} = 8x^2.$

    Thus the denominator behaves like $8x^2$ for small $x$.

    ---

    2. Expand $\sin 5x$

    $\sin 5x \sim 5x \quad \text{as } x \to 0.$

    So the numerator contains a factor $5x$ from the sine.

    ---

    3. Expand $3^{2x}$

    Write $3^{2x} = e^{2x \log 3}$. Using $e^t \sim 1 + t + \frac{t^2}{2}$ for small $t$,

    $3^{2x} \sim 1 + (2\log 3)x + \frac{(2\log 3)^2}{2} x^2 = 1 + 2\log 3 \cdot x + 2(\log 3)^2 x^2.$

    ---

    4. Expand $\sqrt{x+1}$

    $\sqrt{1+x} = (1+x)^{1/2} \sim 1 + \frac{1}{2}x - \frac{1}{8}x^2.$

    ---

    5. Combine the difference

    $3^{2x} - \sqrt{x+1} \sim \left(1 + 2\log 3 \cdot x + 2(\log 3)^2 x^2\right) - \left(1 + \frac12 x - \frac18 x^2\right).$

    Cancel the $1$'s:

    $\sim \left(2\log 3 - \frac12\right)x + \left(2(\log 3)^2 + \frac18\right)x^2.$

    The constant term and $x$ term are present; the $x^2$ term will be of higher order when multiplied by the sine factor.

    ---

    6. Assemble the numerator

    The numerator is

    $(3^{2x} - \sqrt{x+1}) \cdot \sin 5x \sim \left[ \left(2\log 3 - \frac12\right)x + O(x^2) \right] \cdot (5x).$

    So the leading term is

    $5\left(2\log 3 - \frac12\right) x^2.$

    ---

    7. Form the limit

    $L = \lim_{x\to 0} \frac{5\left(2\log 3 - \frac12\right) x^2}{8x^2} = \frac{5}{8}\left(2\log 3 - \frac12\right).$

    Simplify:

    $\frac{5}{8} \cdot 2\log 3 = \frac{5}{4}\log 3, \quad \frac{5}{8} \cdot \left(-\frac12\right) = -\frac{5}{16}.$

    So

    $L = \frac{5}{4}\log 3 - \frac{5}{16}.$

    ---

    8. Match to the options

    Factor $\frac{5}{16}$:

    $L = \frac{5}{16} (4\log 3 - 1).$

    Now $4\log 3 = \log(3^4) = \log 81$, so

    $L = \frac{5}{16} (\log 81 - 1) = \frac{5}{16} \log\left(\frac{81}{e}\right).$

    This is exactly option (B).

    Watch out

    A common mistake is to stop after simplifying the difference $3^{2x} - \sqrt{x+1}$ to only the linear term but forget that the sine contributes another factor of $x$, making the numerator order $x^2$ — matching the denominator's order. Always check the leading power.

    Tip

    Notice that $\log 81 = 4\log 3$, so the answer can be written compactly as $\frac{5}{16}\log(81/e)$. This form appears directly in option (B).

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. 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).