Q.Differentiate w.r.t. $x$: $\cos\left(\tan\sqrt{x + 1}\right)$.

Showing the 12 most recent of 26 on this concept.

1. 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 $\tan^{-1}$. …

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2. 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 $\sec^{2}y=1+\tan^{2}y$: …

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3. 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 …

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4. 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:

   $$$$

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5. 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: …

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6. 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: $$$$

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7. 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: …

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8. 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 $t'=4x-6=2(2x-3)$, …

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9. 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:

   $$$$

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10. 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: $$$$

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11. 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 …

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12. 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. …

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