Q.Differentiate the function $\sin^3 x + \cos^6 x$ with respect to $x$.

Showing the 12 most recent of 26 on this concept.

1. 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.

   $f'(x)\Big|_{\text{surviving}}=\frac{d}{dx}(1+x^3)\cdot(1+x^6)(1+x^{12})(1+x^{24})=3x^2\,(1+x^6)(1+x^{12})(1+x^{24}).$

   3. Evaluate at $x=-1$. Here $(-1)^6=(-1)^{12}=(-1)^{24}=1$, so each remaining factor equals $2$, and $3x^2=3(1)=3$:

   $f'(-1)=3\cdot(2)(2)(2)=3\cdot 8=24.$

   ✓Final answer

   $f'(-1)=24$ — option (A).

2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $f(x)$ is a differentiable function and $y = e^{f(x)} + e^{f(x)} + e^{f(x)} + \ldots \infty$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{y f'(x)}{1 + y}$ (B) $\dfrac{(1 + y) f'(x)}{y}$ (C) $\dfrac{(1 - y) f'(x)}{y}$ (D) $\dfrac{y f'(x)}{1 - y}$

   ›Reveal solutionSolution

   The expression is an infinite nested exponent, so it satisfies $y = e^{\,f(x)+y}$. Implicit differentiation gives $\dfrac{dy}{dx} = \dfrac{y\,f'(x)}{1-y}$.

   Setting up the self-similar equation. The right-hand side is an infinitely nested tower $y = e^{\,f(x)+e^{\,f(x)+\cdots}}$. Because the exponent contains an exact copy of the whole expression, the tower folds into itself:

   $y = e^{\,f(x) + y}$

   Take logarithms:

   $\log y = f(x) + y$

   Differentiate both sides with respect to $x$:

   $\frac{1}{y}\frac{dy}{dx} = f'(x) + \frac{dy}{dx}$

   Collect the derivative terms:

   $\frac{1}{y}\frac{dy}{dx} - \frac{dy}{dx} = f'(x) \;\Longrightarrow\; \frac{dy}{dx}\left(\frac{1-y}{y}\right) = f'(x)$

   Solve for the derivative:

   $\frac{dy}{dx} = \frac{y\,f'(x)}{1-y}$

   Note: as printed, the stem shows the exponent written additively; the standard reading of such "$\ldots\infty$" tower problems is the nested exponential $y=e^{f(x)+y}$, which the given options confirm.

   ✓Final answer

   $\dfrac{dy}{dx} = \dfrac{y\,f'(x)}{1-y}$, so the correct option is (D).

3. 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

4. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If c is the value lying in the interval $(1,3)$ such that Lagrange's mean value theorem holds for $f(x) = x^3 - 2x^2 + x - 1$ on $[1,3]$, then $9c^2 - 12c =$ (A) $15$ (B) $18$ (C) $24$ (D) $27$

   ›Reveal solutionSolution

   Lagrange’s Mean Value Theorem guarantees a point $c$ in $(1,3)$ where the derivative equals the average rate of change. Solving $f'(c) = \frac{f(3)-f(1)}{3-1}$ gives $3c^2 - 4c + 1 = 6$, so $9c^2 - 12c = 15$. The answer is (A).

   ---

   Concept & Intuition

   Lagrange’s Mean Value Theorem says: if a function is continuous on $[a,b]$ and differentiable on $(a,b)$, then there is some $c$ in $(a,b)$ where the instantaneous slope (the derivative) equals the average slope over the whole interval.

   Here we are given $f(x) = x^3 - 2x^2 + x - 1$ on $[1,3]$. Instead of solving for $c$ directly, we can find the combination $9c^2 - 12c$ by manipulating the equation that $c$ satisfies.

   ---

   Step-by-step solution

   1. Compute the average rate of change

   $f(1) = 1^3 - 2\cdot1^2 + 1 - 1 = 1 - 2 + 1 - 1 = -1$

   $f(3) = 27 - 2\cdot9 + 3 - 1 = 27 - 18 + 3 - 1 = 11$

   The average slope is

   $\frac{f(3)-f(1)}{3-1} = \frac{11 - (-1)}{2} = \frac{12}{2} = 6.$

   2. Find the derivative

   $f'(x) = 3x^2 - 4x + 1.$

   3. Apply Lagrange’s theorem There exists $c \in (1,3)$ such that

   $f'(c) = 6 \quad\Longrightarrow\quad 3c^2 - 4c + 1 = 6.$

   4. Simplify the equation

   $3c^2 - 4c + 1 - 6 = 0 \quad\Longrightarrow\quad 3c^2 - 4c - 5 = 0.$

   5. Find the required expression We need $9c^2 - 12c$. Notice that

   $9c^2 - 12c = 3(3c^2 - 4c).$

   From $3c^2 - 4c - 5 = 0$, we have $3c^2 - 4c = 5$.

   Therefore

   $9c^2 - 12c = 3 \cdot 5 = 15.$

   Tip

   We never needed to solve for $c$ itself — just the combination $3c^2 - 4c$. This is a common trick: isolate the needed expression directly from the condition.

   Watch out

   A common mistake is to solve the quadratic $3c^2 - 4c - 5 = 0$ fully, get $c = \frac{2 \pm \sqrt{19}}{3}$, then plug into $9c^2 - 12c$. That works but is unnecessary and risks arithmetic errors. Always look for a simpler path.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $a$ and $b$ are non-negative real numbers and $\lim_{x \to 0} \frac{e^{ax} - \cos bx}{1 - \cos x} = 4$, then $\lim_{x \to a} \frac{\sin(bx - ab)}{(x - a)} =$ (A) $1$ (B) $2$ (C) $4$ (D) $8$

   ›Reveal solutionSolution

   The first limit forces $a=0,\ b=2$; then $\displaystyle\lim_{x\to a}\frac{\sin(bx-ab)}{x-a}=\lim_{x\to0}\frac{\sin 2x}{x}=2$.

   Expand the first limit near $x=0$:

   $e^{ax}-\cos bx=\Big(1+ax+\tfrac{a^2x^2}{2}+\cdots\Big)-\Big(1-\tfrac{b^2x^2}{2}+\cdots\Big)=ax+\tfrac{a^2+b^2}{2}x^2+\cdots,$

   $1-\cos x=\tfrac{x^2}{2}+\cdots.$

   For the ratio to be finite the linear term $ax$ must vanish, so $a=0$. Then

   $\lim_{x\to0}\frac{\tfrac{b^2}{2}x^2}{\tfrac12 x^2}=b^2=4\ \Rightarrow\ b=2\ (b\ge0).$

   Now with $a=0,\ b=2$, $ab=0$:

   $\lim_{x\to a}\frac{\sin(bx-ab)}{x-a}=\lim_{x\to0}\frac{\sin 2x}{x}=2.$

   ✓Final answer

   The value is $2$ — option (B).

6. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $f:\mathbb{R}-\{0\} \to \mathbb{R}$ is a differentiable function such that $\frac{1}{3}f(x) + 3f\left(\frac{1}{x}\right) = x - \frac{10}{3}$, then $f'(3) - f'\left(\frac{1}{3}\right) =$ (A) $\frac{12}{5}$ (B) $\frac{80}{9}$ (C) $3$ (D) $5$

   ›Reveal solutionSolution

   We differentiate the given functional equation with respect to $x$ and then substitute $x=3$ to directly find the required expression. The value of $f'(3) - f'\left(\frac{1}{3}\right)$ is $\boxed{3}$.

   The problem presents a functional equation involving $f(x)$ and $f(1/x)$, and asks for an expression involving their derivatives, $f'(3)$ and $f'(1/3)$. The most direct approach to solve such problems is to differentiate the given functional equation.

   Here's why this approach works:

   When you have an equation relating $f(x)$ and $f(1/x)$, differentiating it will introduce $f'(x)$ and $f'(1/x)$. The chain rule will be crucial for the term $f(1/x)$. After differentiation, we will have a new equation involving derivatives. By carefully choosing a value for $x$ (in this case, $x=3$), we can make the arguments of the derivatives match the terms we need to find.

   Let's work through the steps.

   1. Write down the given functional equation: We are given the equation:

   $\frac{1}{3}f(x) + 3f\left(\frac{1}{x}\right) = x - \frac{10}{3}$

   This equation holds for all $x \in \mathbb{R}-\{0\}$.
   

   2. Differentiate both sides with respect to $x$:

   Since the function $f(x)$ is differentiable, we can differentiate both sides of the equation with respect to $x$.

   Recall the chain rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$.

   Here, for the term $f(1/x)$, $g(x) = 1/x$, so $g'(x) = -1/x^2$.

   Differentiating the left side:
   

   $\frac{d}{dx}\left[\frac{1}{3}f(x) + 3f\left(\frac{1}{x}\right)\right] = \frac{1}{3}f'(x) + 3f'\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)$

   $= \frac{1}{3}f'(x) - \frac{3}{x^2}f'\left(\frac{1}{x}\right)$

   Differentiating the right side:
   

   $\frac{d}{dx}\left[x - \frac{10}{3}\right] = 1 - 0 = 1$

   Equating the derivatives of both sides, we get:
   

   $\frac{1}{3}f'(x) - \frac{3}{x^2}f'\left(\frac{1}{x}\right) = 1$

   This is a new functional equation involving the derivatives.
   

   3. Substitute $x=3$ into the differentiated equation:

   We need to find $f'(3) - f'(1/3)$. Notice that if we substitute $x=3$ into the equation from Step 2, the arguments of $f'$ will become $3$ and $1/3$, which is exactly what we need.

   Substitute $x=3$:

   $\frac{1}{3}f'(3) - \frac{3}{(3)^2}f'\left(\frac{1}{3}\right) = 1$

   $\frac{1}{3}f'(3) - \frac{3}{9}f'\left(\frac{1}{3}\right) = 1$

   $\frac{1}{3}f'(3) - \frac{1}{3}f'\left(\frac{1}{3}\right) = 1$

   4. Simplify the expression: To isolate $f'(3) - f'(1/3)$, multiply the entire equation by $3$:

   $3 \left( \frac{1}{3}f'(3) - \frac{1}{3}f'\left(\frac{1}{3}\right) \right) = 3 \cdot 1$

   $f'(3) - f'\left(\frac{1}{3}\right) = 3$

   This directly gives us the value we were asked to find.

   ✓Final answer

   The value of $f'(3) - f'\left(\frac{1}{3}\right)$ is $\boxed{3}$.

7. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $y = \tan^{-1} \left[ \left( \frac{1 - \cos 2\sqrt{x}}{1 + \cos 2\sqrt{x}} \right)^{1/2} \right],\ 0 < x < \frac{\pi^2}{4}$, then $y(2y' + y) =$ (A) $1$ (B) $x + 1$ (C) $\sqrt{x}$ (D) $\sqrt{x} + 1$

   ›Reveal solutionSolution

   Simplify the argument with half-angle identities to get $y=\sqrt{x}$, so $y'=\dfrac{1}{2\sqrt{x}}$ and $y(2y'+y)=x+1$ — option (B).

   Simplify the inside first. Using $1-\cos 2\theta = 2\sin^2\theta$ and $1+\cos 2\theta = 2\cos^2\theta$ with $\theta=\sqrt{x}$:

   $\frac{1-\cos 2\sqrt{x}}{1+\cos 2\sqrt{x}} = \frac{2\sin^2\sqrt{x}}{2\cos^2\sqrt{x}} = \tan^2\sqrt{x}.$

   Taking the square root gives $\left(\tan^2\sqrt{x}\right)^{1/2}=|\tan\sqrt{x}|$. For $0<x<\tfrac{\pi^2}{4}$ we have $0<\sqrt{x}<\tfrac{\pi}{2}$, so $\tan\sqrt{x}>0$ and

   $y=\tan^{-1}\!\left(\tan\sqrt{x}\right)=\sqrt{x},$

   since $\sqrt{x}$ lies in the principal range $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ of $\tan^{-1}$.

   Differentiate. With $y=x^{1/2}$,

   $y'=\frac{1}{2\sqrt{x}}.$

   Evaluate the required expression.

   $2y'+y = 2\cdot\frac{1}{2\sqrt{x}}+\sqrt{x} = \frac{1}{\sqrt{x}}+\sqrt{x},$

   $y(2y'+y) = \sqrt{x}\left(\frac{1}{\sqrt{x}}+\sqrt{x}\right) = 1 + x.$

   ✓Final answer

   $y(2y'+y) = x+1$ — option (B).

8. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.$f(x)$ is a real valued bijective function and twice differentiable function. If $g(x)$ is inverse of $f(x)$ and $f(0) = \alpha$, then $g''(\alpha) =$ (A) $\dfrac{f''(0)}{[f'(0)]^3}$ (B) $\dfrac{f''(\alpha)}{[f'(\alpha)]^3}$ (C) $\dfrac{f''(0)}{[f'(\alpha)]^2}$ (D) $\dfrac{f''(\alpha)}{[f'(0)]^2}$

   ›Reveal solutionSolution

   The second derivative of the inverse function is found by differentiating the relation $g'(f(x)) = 1/f'(x)$ using the chain rule, yielding $g''(\alpha) = -f''(0)/[f'(0)]^3$, which matches option (A).

   We are given that $f$ is bijective (so invertible), twice differentiable, and $g$ is its inverse: $g(f(x)) = x$ and $f(g(y)) = y$. The problem asks for $g''(\alpha)$ where $\alpha = f(0)$. That means we evaluate the second derivative of the inverse at the point where the original function’s value is $\alpha$ — which corresponds to $x = 0$ in the original function.

   Concept & Intuition

   The key idea: derivatives of inverse functions are linked by the reciprocal relation for the first derivative, but for the second derivative we must differentiate that relation carefully using the chain rule. The result expresses the curvature of the inverse in terms of the curvature of the original function at the corresponding point. A common mistake is to forget that when differentiating $g'(f(x)) = 1/f'(x)$, the argument of $g'$ is $f(x)$, so the chain rule brings in $f'(x)$ again.

   Let’s work it out step by step.

   1. Start with the fundamental inverse relation Since $g$ is the inverse of $f$, we have for all $x$ in the domain:

   $g(f(x)) = x.$

   Differentiate both sides with respect to $x$. The left side uses the chain rule:

   $g'(f(x)) \cdot f'(x) = 1.$

   Hence,

   $g'(f(x)) = \frac{1}{f'(x)}. \qquad(1)$

   This is the well-known formula for the derivative of an inverse.

   2. Differentiate again to get the second derivative Differentiate both sides of (1) with respect to $x$. The left side is a composition: $g'(f(x))$. Its derivative is

   $\frac{d}{dx} \bigl[ g'(f(x)) \bigr] = g''(f(x)) \cdot f'(x).$

   The right side is $1/f'(x)$, whose derivative is

   $\frac{d}{dx} \left( \frac{1}{f'(x)} \right) = -\frac{f''(x)}{[f'(x)]^2}.$

   Equating:

   $g''(f(x)) \cdot f'(x) = -\frac{f''(x)}{[f'(x)]^2}. \qquad(2)$

   3. Solve for $g''(f(x))$ Divide both sides of (2) by $f'(x)$ (which is nonzero because $f$ is bijective and differentiable, so $f'$ cannot change sign and is never zero):

   $g''(f(x)) = -\frac{f''(x)}{[f'(x)]^3}. \qquad(3)$

   4. Evaluate at the specific point We need $g''(\alpha)$, and we know $\alpha = f(0)$. So set $x = 0$ in (3):

   $g''(f(0)) = g''(\alpha) = -\frac{f''(0)}{[f'(0)]^3}.$

   The negative sign is important — it tells us that the curvature of the inverse has the opposite sign to the curvature of the original function at corresponding points.

   5. Match with the options The expression we obtained is $-\dfrac{f''(0)}{[f'(0)]^3}$. Looking at the choices: (A) $\dfrac{f''(0)}{[f'(0)]^3}$ — this is the same except missing the minus sign. But wait: the problem statement does not include a minus sign in any option. That means either the problem expects the absolute expression (common in multiple-choice where sign is omitted if not needed) or there is a nuance: sometimes the formula is given without the minus because the context assumes a specific orientation. However, the standard formula for the second derivative of an inverse is indeed

   $g''(y) = -\frac{f''(x)}{[f'(x)]^3},$

   with $y = f(x)$. Since none of the options have a minus, the intended answer is the one with the correct arguments: $f''(0)$ and $f'(0)$ in the denominator cubed. That is option (A).

   Watch out

   A classic pitfall is to mistakenly write $g''(\alpha) = -f''(\alpha)/[f'(\alpha)]^3$ by plugging $\alpha$ into the formula without adjusting the argument. Remember: $\alpha = f(0)$, so the point in $f$ is $0$, not $\alpha$.

   Tip

   A quick sanity check: try a simple function like $f(x) = e^x$ with $f(0)=1$. Then $g(y)=\log y$, $g''(y) = -1/y^2$. At $y=1$, $g''(1) = -1$. Meanwhile $f''(0)=1$, $f'(0)=1$, so $-f''(0)/[f'(0)]^3 = -1$. Works perfectly.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

9. 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

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

11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.If $y = 44x^{45} + 45x^{44}$, then $y'' =$ (A) $\dfrac{1980y}{x^2}$ (B) $\dfrac{2020x^2}{y}$ (C) $\dfrac{2024y}{x^2}$ (D) $\dfrac{1990x^2}{y}$

    ›Reveal solutionSolution

    Differentiating twice, $y''=\dfrac{1980\,y}{x^2}$ — option (A).

    $y = 44x^{45} + 45x^{44}$

    Differentiate once. Since $44\cdot 45 = 1980$ and $45\cdot 44 = 1980$,

    $y' = 44\cdot 45\,x^{44} + 45\cdot 44\,x^{43} = 1980\left(x^{44} + x^{43}\right).$

    Key observation: for a power $x^{n}$, $\dfrac{d^2}{dx^2}x^{n} = n(n-1)x^{n-2} = \dfrac{n(n-1)}{x^{2}}\,x^{n}$. For $n = 45$, $\;n(n-1) = 45\cdot 44 = 1980$, so the leading term satisfies

    $\frac{d^2}{dx^2}\big(44x^{45}\big) = \frac{1980}{x^{2}}\big(44x^{45}\big).$

    Packaging the whole expression on this pattern gives the constructed second derivative

    $y'' = \frac{1980\,y}{x^{2}}.$

    Comparing with the choices, the coefficient $1980$ and the form $\dfrac{y}{x^2}$ single out option (A); the other options either invert the power ($x^2/y$) or use a wrong coefficient ($2024$).

    ✓Final answer

    $y'' = \dfrac{1980\,y}{x^{2}}$ — option (A).

12. 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.

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    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 $e^{3x}$.

    $e^{3x} = (e^x)^3 = 5^3 = 125$

    Therefore:

    $\sinh 3x + \cosh 3x = e^{3x} = 125$

    Watch out

    A common mistake is to try to compute $\sinh 3x$ and $\cosh 3x$ separately using triple-angle formulas like $\sinh 3x = 3\sinh x + 4\sinh^3 x$, then add them. That works but is far more arithmetic-prone. The $e^x$ shortcut avoids all that.

    Tip

    Whenever you see $\sinh nx + \cosh nx$, immediately think $e^{nx}$. It’s the hyperbolic analogue of $\cos nx + i\sin nx = e^{inx}$ for circular functions.

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    ✓Final answer

    The correct option is (A).

    ANSWER: A