Q.$x$ and $y$ are the sides of two squares such that $y = x - x^2$. Find the rate of change of the area of the second square with respect to the area of the first square.

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.A point P is moving on the curve $x^3 y^4 = 27$. The x-coordinate of P is decreasing at the rate of 8 units per second. When the point P is at (2,2), the y-coordinate of P (A) increases at the rate of 6 units per second (B) decreases at the rate of 6 units per second (C) increases at the rate of 4 units per second (D) decreases at the rate of 4 units per second

   ›Reveal solutionSolution

   Using implicit differentiation with respect to time, we relate the rates of change of $x$ and $y$ on the curve $x^3 y^4 = 27$. Given $\frac{dx}{dt} = -8$ at $(2,2)$, we find $\frac{dy}{dt} = 6$, so the y-coordinate increases at 6 units per second. The correct option is (A).

   We are told that a point $P$ moves along the curve $x^3 y^4 = 27$. The $x$-coordinate is decreasing at 8 units per second, so $\frac{dx}{dt} = -8$. We need the rate of change of the $y$-coordinate, $\frac{dy}{dt}$, at the instant when $P$ is at $(2,2)$.

   Concept and intuition:

   When two variables are linked by an equation, their rates of change are also linked. Differentiating the equation with respect to time $t$ (using the chain rule) gives a relationship between $\frac{dx}{dt}$ and $\frac{dy}{dt}$. This is called related rates. We plug in the known values to solve for the unknown rate.

   1. Start with the given relation:

   $x^3 y^4 = 27$

   Both $x$ and $y$ are functions of time $t$.

   2. Differentiate both sides with respect to $t$: Use the product rule and chain rule.

   $\frac{d}{dt}(x^3 y^4) = \frac{d}{dt}(27)$

   The right side is constant, so its derivative is 0.

   For the left side:

   $\frac{d}{dt}(x^3) \cdot y^4 + x^3 \cdot \frac{d}{dt}(y^4) = 0$

   Now $\frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt}$ and $\frac{d}{dt}(y^4) = 4y^3 \frac{dy}{dt}$.

   So we have:

   $3x^2 \frac{dx}{dt} \cdot y^4 + x^3 \cdot 4y^3 \frac{dy}{dt} = 0$

   3. Simplify the equation:

   $3x^2 y^4 \frac{dx}{dt} + 4x^3 y^3 \frac{dy}{dt} = 0$

   4. Solve for $\frac{dy}{dt}$: Isolate the term with $\frac{dy}{dt}$:

   $4x^3 y^3 \frac{dy}{dt} = -3x^2 y^4 \frac{dx}{dt}$

   Divide both sides by $4x^3 y^3$ (valid since $x$ and $y$ are nonzero at $(2,2)$):

   $\frac{dy}{dt} = -\frac{3x^2 y^4}{4x^3 y^3} \frac{dx}{dt}$

   Simplify the fraction:

   $$$$

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2. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $x = \dfrac{9t^2}{1 + t^4}$ and $y = \dfrac{16t^2}{1 - t^4}$ then $\dfrac{dy}{dx} =$ (A) $\dfrac{16}{9} \left( \dfrac{1 - t^4}{1 + t^4} \right)^3$ (B) $\dfrac{16(1 - t^4)}{9(1 + t^4)}$ (C) $\dfrac{9(1 - t^4)}{16(1 + t^4)}$ (D) $\dfrac{16}{9} \left( \dfrac{1 + t^4}{1 - t^4} \right)^3$

   ›Reveal solutionSolution

   We compute $\frac{dy}{dx}$ by the parametric formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, simplify each derivative, and find that the result matches option (A).

   We are given $x$ and $y$ as functions of a parameter $t$:

   $x = \frac{9t^2}{1 + t^4}, \qquad y = \frac{16t^2}{1 - t^4}.$

   To find $\frac{dy}{dx}$, we use the chain rule for parametric equations:

   $\frac{dy}{dx} = \frac{dy/dt}{dx/dt},$

   provided $dx/dt \neq 0$. This is the natural approach: differentiate each with respect to $t$, then divide.

   ---

   1. Differentiate $x$ with respect to $t$. $x = 9t^2 (1 + t^4)^{-1}$. Use the product rule (or quotient rule):

   $\frac{dx}{dt} = 9 \left[ 2t \cdot (1 + t^4)^{-1} + t^2 \cdot (-1)(1 + t^4)^{-2} \cdot 4t^3 \right].$

   Simplify:

   $\frac{dx}{dt} = 9 \left[ \frac{2t}{1 + t^4} - \frac{4t^5}{(1 + t^4)^2} \right].$

   Put over a common denominator $(1 + t^4)^2$:

   $\frac{dx}{dt} = 9 \cdot \frac{2t(1 + t^4) - 4t^5}{(1 + t^4)^2} = 9 \cdot \frac{2t + 2t^5 - 4t^5}{(1 + t^4)^2} = 9 \cdot \frac{2t - 2t^5}{(1 + t^4)^2}.$

   Factor $2t$:

   $\frac{dx}{dt} = 9 \cdot \frac{2t(1 - t^4)}{(1 + t^4)^2} = \frac{18t(1 - t^4)}{(1 + t^4)^2}.$

   2. Differentiate $y$ with respect to $t$. $y = 16t^2 (1 - t^4)^{-1}$. Similarly:

   $\frac{dy}{dt} = 16 \left[ 2t \cdot (1 - t^4)^{-1} + t^2 \cdot (-1)(1 - t^4)^{-2} \cdot (-4t^3) \right].$

   Notice the minus signs: the derivative of $(1 - t^4)$ is $-4t^3$, so the chain rule gives $(-1)(-4t^3) = +4t^3$. Thus:

   $\frac{dy}{dt} = 16 \left[ \frac{2t}{1 - t^4} + \frac{4t^5}{(1 - t^4)^2} \right].$

   Common denominator $(1 - t^4)^2$:

   $\frac{dy}{dt} = 16 \cdot \frac{2t(1 - t^4) + 4t^5}{(1 - t^4)^2} = 16 \cdot \frac{2t - 2t^5 + 4t^5}{(1 - t^4)^2} = 16 \cdot \frac{2t + 2t^5}{(1 - t^4)^2}.$

   Factor $2t$:

   $\frac{dy}{dt} = 16 \cdot \frac{2t(1 + t^4)}{(1 - t^4)^2} = \frac{32t(1 + t^4)}{(1 - t^4)^2}.$

   3. Form the ratio $\frac{dy}{dx}$.

   $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{ \dfrac{32t(1 + t^4)}{(1 - t^4)^2} }{ \dfrac{18t(1 - t^4)}{(1 + t^4)^2} }.$

   Cancel $t$ (assuming $t \neq 0$; the formula still holds in the limit):

   $\frac{dy}{dx} = \frac{32}{18} \cdot \frac{(1 + t^4)}{(1 - t^4)^2} \cdot \frac{(1 + t^4)^2}{(1 - t^4)}.$

   Simplify $\frac{32}{18} = \frac{16}{9}$. Combine powers:

   $\frac{dy}{dx} = \frac{16}{9} \cdot \frac{(1 + t^4)^3}{(1 - t^4)^3} = \frac{16}{9} \left( \frac{1 + t^4}{1 - t^4} \right)^3.$

   Watch out

   A common mistake is to forget the sign when differentiating $1 - t^4$ or to misplace the powers when simplifying the compound fraction. Double-check the algebra carefully. …

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3. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.The value of $\dfrac{d}{dx} \left[ \log \left( \sin \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right) \right) \right]$ when $x = \sqrt{2}$, is (A) $\dfrac{\sqrt{2} \cot \left( \dfrac{\sqrt{3}}{2} \right)}{6\sqrt{3}}$ (B) $\dfrac{\sqrt{2} \tan \left( \dfrac{\sqrt{3}}{2} \right)}{6\sqrt{3}}$ (C) $\dfrac{\sqrt{2} \cot \left( \dfrac{\sqrt{3}}{2} \right)}{8\sqrt{3}}$ (D) $\dfrac{\sqrt{2} \tan \left( \dfrac{\sqrt{3}}{2} \right)}{8\sqrt{3}}$

   ›Reveal solutionSolution

   To find the derivative of the nested logarithmic function, we apply the chain rule repeatedly from the outermost function inwards. After differentiating, we substitute $x=\sqrt{2}$ and simplify the expression to find the value $\boxed{\dfrac{\sqrt{2} \cot \left( \dfrac{\sqrt{3}}{2} \right)}{8\sqrt{3}}}$.

   The problem asks us to find the derivative of a complex, nested function and then evaluate it at a specific point. The core concept here is the chain rule for differentiation, which allows us to differentiate composite functions. When functions are nested, like $f(g(h(x)))$, the chain rule is applied layer by layer, starting from the outermost function and working inwards.

   If $y = f(u)$ and $u = g(x)$, then the chain rule states:

   $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

   For multiple nested functions, say $y = f(g(h(x)))$, it extends to:

   $\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$

   Let the given function be $y = \log \left( \sin \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right) \right)$. We will apply the chain rule by identifying the layers of functions:

   1. The outermost function is $\log(u)$.
   2. The next layer is $\sin(v)$.
   3. The next layer is $\sqrt{w}$.
   4. The innermost function is $\dfrac{x^2+1}{x^2+2}$.

   Let's differentiate step-by-step:

   1. Differentiate the outermost function (logarithm): The derivative of $\log(f(x))$ is $\frac{1}{f(x)} \cdot f'(x)$. Here, $f(x) = \sin \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right)$. So, the first part of the derivative is:

   $\frac{d}{dx} \left[ \log \left( \sin \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right) \right) \right] = \frac{1}{\sin \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right)} \cdot \frac{d}{dx} \left[ \sin \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right) \right]$

   2. Differentiate the next layer (sine function): Now we need to differentiate $\sin(g(x))$, where $g(x) = \sqrt{\dfrac{x^2+1}{x^2+2}}$. The derivative of $\sin(g(x))$ is $\cos(g(x)) \cdot g'(x)$. Substituting this into our expression from Step 1:

   $\frac{dy}{dx} = \frac{1}{\sin \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right)} \cdot \cos \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right) \cdot \frac{d}{dx} \left[ \sqrt{\dfrac{x^2+1}{x^2+2}} \right]$

   We know that $\frac{\cos \theta}{\sin \theta} = \cot \theta$. So, this simplifies to:
   

   $\frac{dy}{dx} = \cot \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right) \cdot \frac{d}{dx} \left[ \sqrt{\dfrac{x^2+1}{x^2+2}} \right]$

   3. Differentiate the next layer (square root function): Next, we differentiate $\sqrt{h(x)}$, where $h(x) = \dfrac{x^2+1}{x^2+2}$. The derivative of $\sqrt{h(x)}$ (or $h(x)^{1/2}$) is $\frac{1}{2\sqrt{h(x)}} \cdot h'(x)$. Substituting this into our expression:

   $\frac{dy}{dx} = \cot \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right) \cdot \frac{1}{2\sqrt{\dfrac{x^2+1}{x^2+2}}} \cdot \frac{d}{dx} \left[ \dfrac{x^2+1}{x^2+2} \right]$

   4. Differentiate the innermost function (rational function): Finally, we differentiate $\dfrac{x^2+1}{x^2+2}$ using the quotient rule.

      The quotient rule states that if $f(x) = \frac{p(x)}{q(x)}$, then $f'(x) = \frac{p'(x)q(x) - p(x)q'(x)}{(q(x))^2}$.

      Let $p(x) = x^2+1$ and $q(x) = x^2+2$.

      Then $p'(x) = 2x$ and $q'(x) = 2x$.

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

   $= \dfrac{2x(x^2+2 - (x^2+1))}{(x^2+2)^2}$

   $= \dfrac{2x(x^2+2 - x^2-1)}{(x^2+2)^2}$

   $= \dfrac{2x(1)}{(x^2+2)^2} = \dfrac{2x}{(x^2+2)^2}$

   5. Combine all parts of the derivative: Now, substitute the result from Step 4 back into the expression from Step 3:

   $\frac{dy}{dx} = \cot \left( \sqrt{\dfrac{x^2+1}{x^2+2}} \right) \cdot \frac{1}{2\sqrt{\dfrac{x^2+1}{x^2+2}}} \cdot \dfrac{2x}{(x^2+2)^2}$

   Simplify the expression: …
   

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the rate of change of the slope of the tangent drawn to the curve $y = x^3 - 2x^2 + 3x - 2$ at the point $(2, 4)$ is $k$ times the rate of change of its abscissa, then $k=$ (A) $2$ (B) $4$ (C) $6$ (D) $8$

   ›Reveal solutionSolution

   The problem asks for the constant $k$ such that the rate of change of the slope of the tangent equals $k$ times the rate of change of the abscissa. This means we need the second derivative of $y$ with respect to $x$, evaluated at $x=2$, which gives $k=8$.

   We are told that the rate of change of the slope of the tangent is $k$ times the rate of change of the abscissa.

   The “slope of the tangent” is just the first derivative $\frac{dy}{dx}$.

   The “rate of change” of that slope with respect to time (or any parameter) is $\frac{d}{dt}\left(\frac{dy}{dx}\right)$.

   The “rate of change of its abscissa” is $\frac{dx}{dt}$.

   The condition is:

   $\frac{d}{dt}\left(\frac{dy}{dx}\right) = k \cdot \frac{dx}{dt}$

   By the chain rule, $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2} \cdot \frac{dx}{dt}$.

   So the equation becomes:

   $\frac{d^2y}{dx^2} \cdot \frac{dx}{dt} = k \cdot \frac{dx}{dt}$

   Assuming $\frac{dx}{dt} \neq 0$ (the abscissa is changing), we cancel it and get:

   $\frac{d^2y}{dx^2} = k$

   Thus, $k$ is simply the second derivative of $y$ with respect to $x$ at the given point.

   Now we compute:

   1. First derivative:

      $y = x^3 - 2x^2 + 3x - 2$

      $\frac{dy}{dx} = 3x^2 - 4x + 3$

   2. Second derivative:

      $\frac{d^2y}{dx^2} = 6x - 4$

   3. Evaluate at $x=2$: …

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5. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.If $y(\cos x)^{\sin x} = (\sin x)^{\sin x}$ then the value of $\dfrac{dy}{dx}$ at $x = \dfrac{\pi}{4}$ is (A) $0$ (B) $1$ (C) $\sqrt{2}$ (D) $\dfrac{\sqrt{3}}{2}$

   ›Reveal solutionSolution

   The relation simplifies to $y=(\tan x)^{\sin x}$; logarithmic differentiation gives $\dfrac{dy}{dx}=\sqrt{2}$ at $x=\dfrac{\pi}{4}$.

   Simplify. From $y(\cos x)^{\sin x}=(\sin x)^{\sin x}$,

   $y=\frac{(\sin x)^{\sin x}}{(\cos x)^{\sin x}}=(\tan x)^{\sin x}.$

   Logarithmic differentiation.

   $\log y=\sin x\,\log(\tan x),$

   $\frac{1}{y}\frac{dy}{dx}=\cos x\,\log(\tan x)+\sin x\cdot\frac{\sec^2x}{\tan x}.$

   Since $\dfrac{\sec^2x}{\tan x}=\dfrac{1}{\sin x\cos x}$, the second term is $\dfrac{1}{\cos x}=\sec x$:

   $\frac{dy}{dx}=y\big(\cos x\,\log(\tan x)+\sec x\big).$ …

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6. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.If $y = \log \left[ \tan \sqrt{\frac{2x-1}{2x+1}} \right],\ x > 0$, then $\left( \frac{dy}{dx} \right)_{x=1} =$ (A) $\frac{4\sqrt{2} \log 2}{9 \sin \left( \frac{2}{\sqrt{3}} \right)}$ (B) $\frac{4\sqrt{3} \log 2}{9 \sin \left( \frac{\sqrt{3}}{2} \right)}$ (C) $\frac{4\sqrt{3} \log 2}{9 \sin \left( \frac{2}{\sqrt{3}} \right)}$ (D) $\frac{4\sqrt{2} \log 2}{9 \sin \left( \frac{\sqrt{3}}{2} \right)}$

   ›Reveal solutionSolution

   Differentiate the nested function with the chain rule, using $\dfrac{d}{dx}\ln(\tan u) = \dfrac{2}{\sin 2u}\,\dfrac{du}{dx}$. Evaluating at $x = 1$ produces the factors $\sqrt3$ and $\sin\!\left(\tfrac{2}{\sqrt3}\right)$, which single out option (C).

   Step 1 — Set up. Let $u = \sqrt{\dfrac{2x-1}{2x+1}}$, so $y = \log(\tan u)$.

   Step 2 — Differentiate the outer layers.

   $\frac{dy}{dx} = \frac{\sec^2 u}{\tan u}\,\frac{du}{dx} = \frac{1}{\sin u\cos u}\,\frac{du}{dx} = \frac{2}{\sin 2u}\,\frac{du}{dx}.$

   Step 3 — Differentiate $u$.

   $\frac{d}{dx}\!\left(\frac{2x-1}{2x+1}\right) = \frac{4}{(2x+1)^2}, \qquad \frac{du}{dx} = \frac{2}{(2x+1)^{3/2}\sqrt{2x-1}}.$

   Step 4 — Combine.

   $\frac{dy}{dx} = \frac{4}{(2x+1)^{3/2}\sqrt{2x-1}\,\sin 2u}.$

   Step 5 — Evaluate at $x = 1$.

   Here $u = \sqrt{\tfrac13} = \tfrac{1}{\sqrt3}$, so $2u = \tfrac{2}{\sqrt3}$; also $(2x+1)^{3/2} = 3\sqrt3$ and $\sqrt{2x-1} = 1$. Hence …

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7. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.For a given function $y = f(x)$, $\delta y$ denotes the actual error in $y$ corresponding to actual error $\delta x$ in $x$ and $dy$ denotes the approximate value of $\delta y$. If $y = f(x) = 2x^2 - 3x + 4$ and $\delta x = 0.02$, then the value of $\delta y - dy$ when $x = 5$ is (A) 0.0008 (B) 0.008 (C) 0.0004 (D) 0.004

   ›Reveal solutionSolution

   For a quadratic, $\delta y-dy=\tfrac12 f''(x)(\delta x)^2=2(0.02)^2=0.0008$.

   Set up the two changes. With $y=f(x)=2x^2-3x+4$:

   • Actual change: $\delta y=f(x+\delta x)-f(x)$.
   • Approximate change (differential): $dy=f'(x)\,\delta x$, where $f'(x)=4x-3$.

   Exact difference. Using the Taylor expansion of a quadratic, $f(x+h)-f(x)=f'(x)h+\tfrac12 f''(x)h^2$ with $f''(x)=4$:

   $\delta y-dy=\big[f'(x)\delta x+\tfrac12 f''(x)(\delta x)^2\big]-f'(x)\delta x=\tfrac12 f''(x)(\delta x)^2.$ …

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8. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.

   [!FORMULA] $\left[ \frac{d}{dx} \left( (\sin x)^{\cos x} \right) \right]_{x = \frac{\pi}{4}} =$

   (A) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{\sqrt{2}+1}{\sqrt{2}}} (1 + \log \sqrt{2})$ (B) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{1}{\sqrt{2}}} (1 + \log \sqrt{2})$ (C) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{1}{\sqrt{2}}} (1 - \log \sqrt{2})$ (D) $\left( \frac{1}{\sqrt{2}} \right)^{\frac{\sqrt{2}+1}{\sqrt{2}}} (1 - \log \sqrt{2})$

   ›Reveal solutionSolution

   Logarithmic differentiation gives $\dfrac{dy}{dx} = (\sin x)^{\cos x}\left(\dfrac{\cos^2 x}{\sin x} - \sin x\,\log(\sin x)\right)$. Evaluating at $x = \pi/4$ yields $\left(\dfrac{1}{\sqrt 2}\right)^{\frac{\sqrt 2 + 1}{\sqrt 2}}(1 + \log\sqrt 2)$, option (A).

   ---

   Concept & intuition

   Because the exponent $\cos x$ is itself a function of $x$, the ordinary power rule does not apply. Use logarithmic differentiation: take $\log$ of both sides, differentiate implicitly, then multiply back by $y$.

   ---

   1. Take logarithms Let $y = (\sin x)^{\cos x}$. Then

   $\log y = \cos x \,\log(\sin x).$

   2. Differentiate implicitly (product rule on the right):

   $\frac{1}{y}\frac{dy}{dx} = -\sin x\,\log(\sin x) + \cos x\cdot\frac{\cos x}{\sin x} = -\sin x\,\log(\sin x) + \frac{\cos^2 x}{\sin x}.$

   3. Solve for the derivative

   $\frac{dy}{dx} = (\sin x)^{\cos x}\left(\frac{\cos^2 x}{\sin x} - \sin x\,\log(\sin x)\right).$

   4. Evaluate at $x = \pi/4$ where $\sin x = \cos x = \dfrac{1}{\sqrt 2}$:

      • Base: $(\sin x)^{\cos x} = \left(\dfrac{1}{\sqrt 2}\right)^{1/\sqrt 2}$.
      • $\dfrac{\cos^2 x}{\sin x} = \dfrac{1/2}{1/\sqrt 2} = \dfrac{1}{\sqrt 2}$.
      • $-\sin x\,\log(\sin x) = -\dfrac{1}{\sqrt 2}\log\!\left(\dfrac{1}{\sqrt 2}\right) = -\dfrac{1}{\sqrt 2}(-\log\sqrt 2) = \dfrac{1}{\sqrt 2}\log\sqrt 2.$

      The bracket becomes

   $\frac{1}{\sqrt 2} + \frac{1}{\sqrt 2}\log\sqrt 2 = \frac{1}{\sqrt 2}\,(1 + \log\sqrt 2).$

   5. Combine $$\left.\frac{dy}{dx}\right|_{\pi/4} = \left(\frac{1}{\sqrt 2}\right)^{1/\sqrt 2}\cdot\frac{1}{\sqrt 2}\,(1 + \log\sqrt 2). …$$

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9. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $y = \tan^{-1} \left( \frac{\sqrt{1+9x^2}-1}{3x} \right)$ then $\left( \frac{dy}{dx} \right)_{x = \frac{1}{3}} =$ (A) $\frac{2}{3}$ (B) $\frac{1}{3}$ (C) $\frac{3}{4}$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The key is to simplify the inverse tangent expression using a trigonometric substitution, turning it into a linear function of $x$, which makes differentiation trivial. The derivative at $x = \frac13$ is $\frac{3}{4}$, so option (C) is correct.

   ---

   The expression inside the arctan looks messy, but the presence of $\sqrt{1+9x^2}$ suggests a substitution like $3x = \tan\theta$ or $3x = \sinh t$. Since we have a square root of a sum of squares, the hyperbolic substitution is cleaner: let $3x = \sinh t$. Then $\sqrt{1+9x^2} = \sqrt{1+\sinh^2 t} = \cosh t$. This turns the fraction into something like $\frac{\cosh t - 1}{\sinh t}$, which simplifies using hyperbolic half-angle identities. The result is that the arctan becomes just $t/2$, i.e., a linear function of $t$, and hence a simple function of $x$.

   ---

   1. Substitute to simplify Let $3x = \sinh t$, so $x = \frac13 \sinh t$. Then

   $\sqrt{1+9x^2} = \sqrt{1+\sinh^2 t} = \cosh t.$

   The argument of the arctan becomes

   $\frac{\sqrt{1+9x^2} - 1}{3x} = \frac{\cosh t - 1}{\sinh t}.$

   2. Use the hyperbolic half-angle identity Recall: $\cosh t - 1 = 2\sinh^2(t/2)$ and $\sinh t = 2\sinh(t/2)\cosh(t/2)$. Hence

   $\frac{\cosh t - 1}{\sinh t} = \frac{2\sinh^2(t/2)}{2\sinh(t/2)\cosh(t/2)} = \frac{\sinh(t/2)}{\cosh(t/2)} = \tanh(t/2).$

   3. Simplify the inverse tangent So

   $y = \tan^{-1}\bigl(\tanh(t/2)\bigr).$

   But $\tanh(t/2)$ is always between $-1$ and $1$, and $\tan^{-1}$ of a hyperbolic tangent is not simply $t/2$ — wait, we need a different identity. Actually, there is a known relation:

   $\tan^{-1}\left(\frac{\cosh t - 1}{\sinh t}\right) = \frac{t}{2}.$

   Let’s verify: $\tan(t/2) = \frac{\sin t}{1+\cos t}$, but here we have hyperbolic functions. The correct identity is:

   $\tanh^{-1} u = \frac12 \log\frac{1+u}{1-u},$

   but we have $\tan^{-1}$, not $\tanh^{-1}$. So let’s check numerically: if $t=1$, $\frac{\cosh 1 - 1}{\sinh 1} \approx 0.462$, $\tan^{-1}(0.462) \approx 0.433$, and $t/2 = 0.5$. Not equal. So the direct half-angle idea needs adjustment.

   Better approach: Use the identity

   $\tan^{-1}\left(\frac{\sqrt{1+a^2}-1}{a}\right) = \frac12 \tan^{-1} a,$

   for $a>0$. Let’s prove it: set $a = \tan\theta$, then $\sqrt{1+a^2} = \sec\theta$, so

   $$$$

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10. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
    Q.If $y = (e^{2x} - 4)(6e^{2x} - 5e^x + 1)$, then $\left(\frac{dy}{dx}\right)_{x=0} - \left(\frac{d^2y}{dx^2}\right)_{x=0} =$ (A) $0$ (B) $-5$ (C) $4$ (D) $6$

    ›Reveal solutionSolution

    Differentiate the product, evaluate the first and second derivatives at $x=0$, and subtract: the result is $4$.

    Let $u = e^{2x}-4$ and $v = 6e^{2x}-5e^{x}+1$, so $y = uv$.

    Derivatives of the factors:

    $u' = 2e^{2x},\quad u'' = 4e^{2x}$

    $v' = 12e^{2x}-5e^{x},\quad v'' = 24e^{2x}-5e^{x}$

    Values at $x=0$ (where $e^{2x}=1,\ e^{x}=1$):

    $u = -3,\ v = 2,\quad u' = 2,\ v' = 7,\quad u'' = 4,\ v'' = 19$

    First derivative:

    $\left(\frac{dy}{dx}\right)_{0} = u'v + uv' = (2)(2) + (-3)(7) = 4 - 21 = -17$

    Second derivative: …

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11. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
    Q.By the application of derivatives, the approximate value of $\sqrt{242}$ is (A) 2.9085 (B) 2.9975 (C) 2.9527 (D) 2.8529

    ›Reveal solutionSolution

    The answer choices (all near 3) show this is the fifth root $\sqrt[5]{242}$, not the square root. Linearising $f(x)=x^{1/5}$ about $x=243$ (where $3^5=243$) gives $\sqrt[5]{242}\approx 2.9975$, option (B).

    This is a linear (tangent-line) approximation: near a point $a$ where the value is known exactly,

    $f(a+\Delta x)\approx f(a)+f'(a)\,\Delta x.$

    Because the options cluster around 3 (whereas $\sqrt{242}\approx 15.6$), the quantity being approximated is $\sqrt[5]{242}$, and the natural reference point is $243=3^5$.

    1. Choose the point. Let $f(x)=x^{1/5}$. Then $f(243)=243^{1/5}=3$ exactly. Take $a=243$ and $\Delta x = 242-243 = -1$.

    2. Derivative. $f'(x)=\tfrac15 x^{-4/5}$. Since $243^{1/5}=3$, we have $243^{4/5}=3^4=81$, so

       $f'(243)=\frac15\cdot\frac{1}{81}=\frac{1}{405}.$ …

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12. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.The radius of a cone of height 9 units is changed from 2 units to 2.12 units. The exact change and approximate change in the volume of the cone are respectively (A) $(1.4437)\pi, (1.44)\pi$ (B) $(1.4832)\pi, (1.479)\pi$ (C) $(1.4842)\pi, (1.48)\pi$ (D) $(1.4832)\pi, (1.44)\pi$

    ›Reveal solutionSolution

    The exact change in volume is found by subtracting the original volume from the new volume; the approximate change uses the derivative (differential) of the volume with respect to the radius. The exact change is $(1.4832)\pi$ and the approximate change is $(1.44)\pi$, so the correct option is (D).

    Concept & Intuition

    The volume of a cone is $V = \frac{1}{3}\pi r^2 h$. Here height $h$ is fixed at 9, so $V = 3\pi r^2$. When the radius changes from $r=2$ to $r=2.12$, the exact change is simply $V(2.12) - V(2)$. The approximate change uses the differential $dV = V'(r)\, dr$, which gives a linear approximation valid for small changes. This is a classic application of calculus: the derivative tells us the instantaneous rate of change, and multiplying by the small change in radius gives a quick estimate.

    1. Write the volume formula

       $V = \frac{1}{3}\pi r^2 h$. With $h=9$, this becomes $V = 3\pi r^2$.

    2. Compute the exact change

       Original volume: $V(2) = 3\pi (2)^2 = 12\pi$.

       New volume: $V(2.12) = 3\pi (2.12)^2 = 3\pi \times 4.4944 = 13.4832\pi$.

       Exact change: $\Delta V = 13.4832\pi - 12\pi = (1.4832)\pi$.

    3. Compute the approximate change using differentials

       Derivative: $\frac{dV}{dr} = 6\pi r$. At $r=2$, $\frac{dV}{dr} = 12\pi$.

       Change in radius: $dr = 2.12 - 2 = 0.12$.

       Approximate change: $dV = 12\pi \times 0.12 = 1.44\pi$. …

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