Q.Find $\frac{dy}{dx}$ in the following: $e^x \sin 5x$

Showing the 12 most recent of 26 on this concept.

1. 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 $-5\cos x$. …

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2. 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 $g(x)$:

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

   *   $\frac{d}{dx} (-1) = 0$
   *   $\frac{d}{dx} (\cot x) = -\csc^2 x$ …
   

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3. 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.

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

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4. 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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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 $t'=4x-6=2(2x-3)$, …

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6. 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: …

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7. 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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8. 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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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 2022Set eng-2022-07-20-AN1 markMCQ
    Q.If $f(x) = \sum_{p=1}^{7} p^2 \sin^{-1} \left( \frac{4}{5} \sin(px) - \frac{3}{5} \cos(px) \right)$ then the value of $\frac{df}{dx}$ at $x=1$ is (Given that $\sin^{-1}(\sin x) = x$) (A) 0 (B) 628 (C) 1140 (D) 784

    ›Reveal solutionSolution

    The core idea is to simplify the argument of the inverse sine function using a trigonometric identity, which then allows us to use the given property $\sin^{-1}(\sin x) = x$. After simplification, the function $f(x)$ becomes a sum of linear terms, making its derivative straightforward to calculate. The final value of $\frac{df}{dx}$ at $x=1$ is $\boxed{784}$.

    The problem asks for the derivative of a function $f(x)$ at a specific point. The function $f(x)$ involves a sum and an inverse trigonometric function whose argument is a linear combination of $\sin(px)$ and $\cos(px)$. The key to solving this problem lies in simplifying the argument of the $\sin^{-1}$ function.

    Concept and Intuition

    1. Trigonometric Transformation: An expression of the form $a \sin \theta + b \cos \theta$ can always be rewritten as a single sine or cosine function. Specifically, we can write $a \sin \theta + b \cos \theta = R \sin(\theta + \alpha)$, where $R = \sqrt{a^2+b^2}$, $\cos \alpha = \frac{a}{R}$, and $\sin \alpha = \frac{b}{R}$. This transformation is crucial because it allows us to simplify the argument of $\sin^{-1}$.
    2. Inverse Sine Property: The problem explicitly states that $\sin^{-1}(\sin x) = x$. This is a very important piece of information. Normally, $\sin^{-1}(\sin x)$ equals $x$ only for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. However, by providing this identity, the problem simplifies the situation, allowing us to directly replace $\sin^{-1}(\sin(\text{expression}))$ with the $\text{expression}$ itself, regardless of its range. This avoids complex principal value considerations.
    3. Differentiation of a Sum: The function $f(x)$ is a sum of terms. The derivative of a sum is the sum of the derivatives, which simplifies the differentiation process.

    Let's apply these concepts step-by-step.

    1. Simplify the argument of $\sin^{-1}$: The argument of the inverse sine function is $\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px)$. This is in the form $a \sin \theta + b \cos \theta$, where $a = \frac{4}{5}$, $b = -\frac{3}{5}$, and $\theta = px$. First, calculate $R = \sqrt{a^2+b^2}$:

    $R = \sqrt{\left(\frac{4}{5}\right)^2 + \left(-\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25} + \frac{9}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1$

    Now, we want to express the argument as $R \sin(\theta - \alpha)$.
    We need $\cos \alpha = \frac{a}{R} = \frac{4/5}{1} = \frac{4}{5}$ and $\sin \alpha = \frac{b}{R} = \frac{-3/5}{1} = -\frac{3}{5}$.
    Let $\alpha_0$ be an angle such that $\cos \alpha_0 = \frac{4}{5}$ and $\sin \alpha_0 = \frac{3}{5}$. (This $\alpha_0$ is a constant acute angle, specifically $\alpha_0 = \tan^{-1}(\frac{3}{4})$).
    Then, the expression becomes:
    

    $1 \cdot \left( \cos \alpha_0 \sin(px) - \sin \alpha_0 \cos(px) \right)$

    Using the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, with $A=px$ and $B=\alpha_0$:
    

    $\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px) = \sin(px - \alpha_0)$

    So, the argument simplifies to $\sin(px - \alpha_0)$.
    

    2. Substitute the simplified argument back into $f(x)$:

    Now, $f(x)$ can be written as:

    $f(x) = \sum_{p=1}^{7} p^2 \sin^{-1} \left( \sin(px - \alpha_0) \right)$

    3. Apply the given identity $\sin^{-1}(\sin x) = x$: …

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11. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.If $\sinh x = \dfrac{12}{5}$, then $\sinh 3x + \cosh 3x =$ (A) $125$ (B) $144$ (C) $169$ (D) $216$

    ›Reveal solutionSolution

    The key is to use the identity $\sinh 3x + \cosh 3x = e^{3x}$, then find $e^x$ from $\sinh x = \frac{12}{5}$ using $\cosh x = \sqrt{1+\sinh^2 x}$ and $e^x = \sinh x + \cosh x$. The result is $125$, so the correct option is (A).

    ---

    The problem asks for $\sinh 3x + \cosh 3x$ given $\sinh x = \frac{12}{5}$. The direct approach would be to compute $\sinh 3x$ and $\cosh 3x$ using triple-angle formulas, but that’s messy. Instead, recall the elegant identity:

    For any real $x$, $\sinh x + \cosh x = e^x$.

    Similarly, $\sinh 3x + \cosh 3x = e^{3x}$.

    So the problem reduces to finding $e^{3x}$ from $\sinh x = \frac{12}{5}$. That’s much simpler.

    ---

    Step-by-step reasoning:

    1. Find $\cosh x$ from $\sinh x$. The fundamental identity for hyperbolic functions is:

    $\cosh^2 x - \sinh^2 x = 1$

    Given $\sinh x = \frac{12}{5}$, we have:

    $\cosh^2 x = 1 + \left(\frac{12}{5}\right)^2 = 1 + \frac{144}{25} = \frac{169}{25}$

    Since $\cosh x \ge 1$ for all real $x$, we take the positive root:

    $\cosh x = \sqrt{\frac{169}{25}} = \frac{13}{5}$

    2. Find $e^x$ using the sum identity. As noted:

    $e^x = \sinh x + \cosh x = \frac{12}{5} + \frac{13}{5} = \frac{25}{5} = 5$

    So $e^x = 5$.

    3. Compute $e^{3x}$. $$$$

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12. 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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