Q.Find $\frac{dy}{dx}$ in the following: $\cos (\sin x)$

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 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).

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 $\sec \left( \log_2 y^2 \right) = \csc \left( \log_2 x^2 \right)$, 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 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $y = e^{2x} + \sin x$, then $2y'' - 5y' + 2y =$ (A) $4 \sin x$ (B) $-5 \cos x$ (C) $-4 \sin x$ (D) $5 \cos x$

   ›Reveal solutionSolution

   The key idea is to compute the first and second derivatives of $y = e^{2x} + \sin x$, substitute them into $2y'' - 5y' + 2y$, and simplify. The result is $-5 \cos x$, which corresponds to option (B).

   We start with the function

   $y = e^{2x} + \sin x.$

   The expression we need is $2y'' - 5y' + 2y$. Instead of solving a differential equation, we simply differentiate and substitute — this is a direct computation.

   1. First derivative Differentiate term by term:

   $y' = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(\sin x) = 2e^{2x} + \cos x.$

   2. Second derivative Differentiate $y'$:

   $y'' = \frac{d}{dx}(2e^{2x}) + \frac{d}{dx}(\cos x) = 4e^{2x} - \sin x.$

   3. Substitute into $2y'' - 5y' + 2y$

   $2y'' = 2(4e^{2x} - \sin x) = 8e^{2x} - 2\sin x,$

   $-5y' = -5(2e^{2x} + \cos x) = -10e^{2x} - 5\cos x,$

   $2y = 2(e^{2x} + \sin x) = 2e^{2x} + 2\sin x.$

   4. Add them together Combine the $e^{2x}$ terms: $8e^{2x} - 10e^{2x} + 2e^{2x} = 0$. Combine the $\sin x$ terms: $-2\sin x + 2\sin x = 0$. The only remaining term is $-5\cos x$.

   Watch out

   A common mistake is forgetting the chain rule when differentiating $e^{2x}$ — it’s $2e^{2x}$, not $e^{2x}$. Also, note that the $e^{2x}$ terms cancel completely, so the answer depends only on the trigonometric part.

   Tip

   If you suspect the $e^{2x}$ part will cancel (since $y = e^{2x}$ satisfies $y'' - 4y = 0$), you can focus only on the $\sin x$ term from the start. Here, $2y'' - 5y' + 2y$ applied to $\sin x$ gives $-5\cos x$ directly.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

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 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

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

10. 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

11. 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

12. 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).