Q.Differentiate w.r.t. $x$: $2^{\cos^2 x}$.

Showing the 12 most recent of 14 on this concept.

1. KCET 2022Set C-41 markMCQ
   Q.If $y = x^{\sin x} + (\sin x)^x$ then $\frac{dy}{dx}$ at $x = \frac{\pi}{2}$ is (A) $\pi \log \frac{\pi}{2}$ (B) 1 (C) $\frac{\pi^2}{2}$ (D) 0

   ›Reveal solutionSolution

   Both terms are of the form (variable)$^{\text{variable}}$, so differentiate each by taking logarithms; at $x=\pi/2$ the substitutions $\sin=1,\ \cos=0,\ \cot=0,\ \log 1=0$ collapse everything to $1+0$.

   Step 1 — Why logarithmic differentiation

   Neither $x^{\sin x}$ nor $(\sin x)^{x}$ is a power function or an exponential function — the base and the exponent both vary. So neither the power rule nor the exponential rule applies directly. The standard tool is to take $\log$, use $\log(a^{b})=b\log a$, and differentiate implicitly. Split the sum:

   $y=u+v,\quad u=x^{\sin x},\quad v=(\sin x)^{x},\qquad \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

   Step 2 — Differentiate $u=x^{\sin x}$

   $\log u=\sin x\,\log x$

   Differentiate both sides (product rule on the right):

   $\frac{1}{u}\frac{du}{dx}=\cos x\,\log x+\sin x\cdot\frac{1}{x}$

   $\frac{du}{dx}=x^{\sin x}\left[\cos x\,\log x+\frac{\sin x}{x}\right]$

   Now put $x=\dfrac{\pi}{2}$, where $\sin x=1$ and $\cos x=0$:

   $\frac{du}{dx}=\left(\frac{\pi}{2}\right)^{1}\left[0\cdot\log\frac{\pi}{2}+\frac{1}{\pi/2}\right]=\frac{\pi}{2}\cdot\frac{2}{\pi}=1$

   Step 3 — Differentiate $v=(\sin x)^{x}$

   $\log v=x\,\log(\sin x)$ …

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2. KCET 2025Set A-11 markMCQ
   Q.The derivative of $\sin x$ with respect to $\log x$ is (A) $\cos x$ (B) $x \cos x$ (C) $\dfrac{\cos x}{\log x}$ (D) $\dfrac{\cos x}{x}$

   ›Reveal solutionSolution

   "Derivative of $u$ with respect to $v$" means $\dfrac{du}{dv}=\dfrac{du/dx}{dv/dx}$ — differentiate both with respect to $x$ and divide.

   Step 1 — Name the two functions.

   $u=\sin x, \qquad v=\log x \quad (x>0).$

   We are asked for $\dfrac{du}{dv}$, not $\dfrac{du}{dx}$.

   Step 2 — The chain rule in ratio form. Since both are functions of the common variable $x$,

   $\frac{du}{dv}=\frac{du}{dx}\cdot\frac{dx}{dv}=\frac{\;du/dx\;}{\;dv/dx\;}\qquad\left(\text{valid where }\frac{dv}{dx}\neq 0\right).$

   Step 3 — Differentiate each with respect to $x$.

   $\frac{du}{dx}=\cos x,\qquad \frac{dv}{dx}=\frac{1}{x}.$

   Step 4 — Divide. …

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3. KCET 2020Set A-11 markMCQ
   Q.If $2^x + 2^y = 2^{x+y}$, then $\frac{dy}{dx}$ is (A) $2^{y-x}$ (B) $-2^{y-x}$ (C) $2^{x-y}$ (D) $\frac{2^y - 1}{2^x - 1}$

   ›Reveal solutionSolution

   $\dfrac{dy}{dx} = -2^{\,y-x}$ — option (B).

   Differentiate $2^x + 2^y = 2^{x+y}$ with respect to $x$ (each term contributes a common factor $\log 2$, which cancels):

   $2^x + 2^y\dfrac{dy}{dx} = 2^{x+y}\left(1 + \dfrac{dy}{dx}\right)$

   Collect $\dfrac{dy}{dx}$:

   $\dfrac{dy}{dx}\left(2^y - 2^{x+y}\right) = 2^{x+y} - 2^x \;\Rightarrow\; \dfrac{dy}{dx} = \dfrac{2^x\left(2^y - 1\right)}{2^y\left(1 - 2^x\right)}$ …

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4. KCET 2018Set A-11 markMCQ
   Q.If $\cos y = x \cos (a + y)$ with $\cos a \neq \pm 1$, then $\frac{dy}{dx}$ is equal to (A) $\frac{\sin a}{\cos^2(a+y)}$ (B) $\frac{\cos^2(a+y)}{\sin a}$ (C) $\frac{\cos a}{\sin^2(a+y)}$ (D) $\frac{\cos^2(a+y)}{\cos a}$

   ›Reveal solutionSolution

   Solve for $x$ explicitly, differentiate $x$ with respect to $y$ (much cleaner than implicit differentiation), then invert.

   Step 1 — Express $x$ explicitly.

   Given $\cos y = x\cos(a+y)$ with $\cos a \neq \pm 1$,

   $x = \frac{\cos y}{\cos(a+y)}$

   Step 2 — Differentiate with respect to $y$ (quotient rule).

   $\frac{dx}{dy} = \frac{\cos(a+y)\cdot(-\sin y) - \cos y\cdot\big(-\sin(a+y)\big)}{\cos^{2}(a+y)}$

   $= \frac{\sin(a+y)\cos y - \cos(a+y)\sin y}{\cos^{2}(a+y)}$

   Step 3 — Collapse the numerator with the sine-difference identity.

   $\sin A\cos B - \cos A \sin B = \sin(A-B)$

   with $A = a+y$, $B = y$:

   $\sin(a+y)\cos y - \cos(a+y)\sin y = \sin\big((a+y)-y\big) = \sin a$

   So

   $\frac{dx}{dy} = \frac{\sin a}{\cos^{2}(a+y)}$ …

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5. KCET 2018Set A-11 markMCQ
   Q.Let $f(x) = x - \frac{1}{x}$ then $f'(-1)$ is (A) 0 (B) 2 (C) 1 (D) $-2$

   ›Reveal solutionSolution

   Rewrite $\frac{1}{x}$ as $x^{-1}$, apply the power rule term by term, then evaluate at $x = -1$.

   Step 1 — Rewrite in power form.

   $f(x) = x - \frac{1}{x} = x - x^{-1}$

   Writing the reciprocal as a negative power lets us use the single power rule $\frac{d}{dx}x^{n} = n x^{\,n-1}$ on both terms.

   Step 2 — Differentiate.

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

   Note the double negative: the derivative of $-x^{-1}$ is $+x^{-2}$. Dropping this sign gives $1 - \frac{1}{x^{2}} = 0$ at $x=-1$ — which is exactly the trap behind option (A).

   Step 3 — Evaluate at $x = -1$.

   $f'(-1) = 1 + \frac{1}{(-1)^{2}} = 1 + \frac{1}{1} = 2$ …

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.If $y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}, \quad$ then $2 x y \frac{d y}{d x}$ is equal to (A) $x+\frac{a}{x}$ (B) $\frac{x^2+a^2}{a x}$ (C) $\frac{x}{a}-\frac{a}{x}$ (D) $\frac{a}{x}-\frac{x}{a}$

   ›Reveal solutionSolution

   The key is to simplify $y$ before differentiating, using the identity $(\sqrt{u} + \sqrt{1/u})^2 = u + 1/u + 2$. This avoids messy chain rules and leads directly to $2xy\frac{dy}{dx} = \frac{x}{a} - \frac{a}{x}$, which matches option (C).

   Concept & Intuition

   When a function involves sums of square roots of reciprocals, squaring it often reveals a simpler algebraic relationship. Here, $y = \sqrt{x/a} + \sqrt{a/x}$ looks symmetric. Instead of differentiating directly (which would involve messy chain rules and square roots), we can square both sides to get a polynomial-like relation. Then implicit differentiation becomes clean and straightforward.

   Step-by-step solution

   1. Square both sides Let $y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}$. Square:

   $y^2 = \frac{x}{a} + \frac{a}{x} + 2\sqrt{\frac{x}{a} \cdot \frac{a}{x}} = \frac{x}{a} + \frac{a}{x} + 2.$

   The cross-term simplifies because $\sqrt{\frac{x}{a} \cdot \frac{a}{x}} = \sqrt{1} = 1$.

   2. Rewrite as an implicit relation So we have:

   $y^2 = \frac{x}{a} + \frac{a}{x} + 2.$

   This is much simpler than the original form.

   3. Differentiate implicitly with respect to $x$ Differentiate both sides:

   $2y \frac{dy}{dx} = \frac{1}{a} - \frac{a}{x^2}.$

   (Recall $\frac{d}{dx}(x^{-1}) = -1/x^2$, so $\frac{d}{dx}(a/x) = -a/x^2$.)

   4. Multiply both sides by $x$ We want $2xy\frac{dy}{dx}$, so multiply the equation by $x$: …

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7. COMEDK 2023Set 2023-E1 markMCQ
   Q.If $f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}$ then the value of $f^{\prime}(0)$ is equal to (A) 15 (B) 2 (C) 13 (D) 11

   ›Reveal solutionSolution

   Therefore f'(0) = f(0) * 13 = 13.

   Concept: logarithmic differentiation for a product/quotient of powers.

   f(x) = (x+1)^7 * sqrt(1 + x^2) / (x^2 - x + 1)^6.

   Take logs (near x = 0 all factors are positive):

   log f = 7 log(x + 1) + (1/2) log(1 + x^2) - 6 log(x^2 - x + 1).

   Differentiate:

   f'/f = 7/(x + 1) + (1/2)(2x)/(1 + x^2) - 6(2x - 1)/(x^2 - x + 1)

   = 7/(x + 1) + x/(1 + x^2) - 6(2x - 1)/(x^2 - x + 1). …

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8. KCET 2023Set A-21 markMCQ
   Q.If $y = a\sin x + b\cos x$, then $y^2 + \left(\dfrac{dy}{dx}\right)^2$ is a (A) function of $y$ (B) function of $x$ and $y$ (C) constant (D) function of $x$

   ›Reveal solutionSolution

   The expression $y^2 + \left(\frac{dy}{dx}\right)^2$ simplifies to a constant $a^2 + b^2$, independent of $x$ and $y$.

   The key insight here is that when you have a linear combination of $\sin x$ and $\cos x$, the derivative simply swaps and alternates signs between them. Squaring and adding the function and its derivative often produces a Pythagorean identity that cancels the $x$-dependence entirely.

   Let’s work through it step by step.

   1. Write down the given function and its derivative.

      We have $y = a\sin x + b\cos x$.

      Differentiating term by term:

      $\frac{dy}{dx} = a\cos x - b\sin x$.

   2. Square both $y$ and $\frac{dy}{dx}$.

      $y^2 = (a\sin x + b\cos x)^2 = a^2\sin^2 x + 2ab\sin x\cos x + b^2\cos^2 x$.

      $\left(\frac{dy}{dx}\right)^2 = (a\cos x - b\sin x)^2 = a^2\cos^2 x - 2ab\sin x\cos x + b^2\sin^2 x$.

   3. Add the two squares.

      $y^2 + \left(\frac{dy}{dx}\right)^2 = (a^2\sin^2 x + b^2\cos^2 x + a^2\cos^2 x + b^2\sin^2 x) + (2ab\sin x\cos x - 2ab\sin x\cos x)$.

      The cross terms cancel exactly. Group the $\sin^2$ and $\cos^2$ terms:

      $= a^2(\sin^2 x + \cos^2 x) + b^2(\cos^2 x + \sin^2 x)$. …

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9. KCET 2025Set A-11 markMCQ
   Q.A function $f(x) = \begin{cases} \dfrac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}, & \text{if } x \neq 0 \\[4pt] 0, & \text{if } x = 0 \end{cases}$ is (A) continuous at $x = 0$ (B) not continuous at $x = 0$ (C) differentiable at $x = 0$ (D) differentiable at $x = 0$, but not continuous at $x = 0$

   ›Reveal solutionSolution

   Evaluate the two one-sided limits: $e^{1/x} \to \infty$ from the right and $\to 0$ from the left, giving limits $+1$ and $-1$ — they disagree, so $f$ is discontinuous at 0.

   1. The continuity test. $f$ is continuous at $x=0$ iff

   $\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) = f(0)$

   Here $f(0) = 0$ by definition. The whole question hinges on the behaviour of the exponent $\tfrac{1}{x}$, which blows up in opposite directions on the two sides of 0.

   2. Right-hand limit ($x \to 0^{+}$). Then $\tfrac1x \to +\infty$, so $t = e^{1/x} \to +\infty$. Divide numerator and denominator by $t$ (the standard trick when a term dominates):

   $\lim_{x\to 0^{+}} \frac{e^{1/x}-1}{e^{1/x}+1} = \lim_{t\to\infty} \frac{t-1}{t+1} = \lim_{t\to\infty} \frac{1 - \tfrac1t}{1 + \tfrac1t} = \frac{1-0}{1+0} = +1$

   3. Left-hand limit ($x \to 0^{-}$). Now $\tfrac1x \to -\infty$, so $t = e^{1/x} \to 0$. Substitute directly:

   $\lim_{x\to 0^{-}} \frac{e^{1/x}-1}{e^{1/x}+1} = \frac{0-1}{0+1} = -1$

   4. Compare.

   $\lim_{x\to 0^{-}} f(x) = -1 \;\neq\; +1 = \lim_{x\to 0^{+}} f(x)$ …

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10. KCET 2018Set A-11 markMCQ
    Q.If $x, y, z \in \mathbb{R}$, then the value of determinant $\begin{vmatrix} (5^x + 5^{-x})^2 & (5^x - 5^{-x})^2 & 1 \\ (6^x + 6^{-x})^2 & (6^x - 6^{-x})^2 & 1 \\ (7^x + 7^{-x})^2 & (7^x - 7^{-x})^2 & 1 \end{vmatrix}$ is (A) $10$ (B) $12$ (C) $1$ (D) $0$

    ›Reveal solutionSolution

    The identity $(a+b)^2-(a-b)^2 = 4ab$ makes column 1 minus column 2 equal to the constant $4$ in every row, so the columns are linearly dependent and the determinant is $0$.

    Step 1 — Write the determinant.

    $\Delta = \begin{vmatrix} (5^x + 5^{-x})^2 & (5^x - 5^{-x})^2 & 1 \\ (6^x + 6^{-x})^2 & (6^x - 6^{-x})^2 & 1 \\ (7^x + 7^{-x})^2 & (7^x - 7^{-x})^2 & 1 \end{vmatrix}$

    Step 2 — Use the algebraic identity row-wise.

    For any base $a>0$, put $u = a^{x}$ and $v = a^{-x}$. Then $uv = a^{x}a^{-x} = a^{0} = 1$, and

    $(u+v)^2 - (u-v)^2 = 4uv = 4 .$

    So for each of the three rows (bases $5$, $6$, $7$ alike) the first entry minus the second entry equals exactly $4$.

    Step 3 — Column operation.

    A determinant is unchanged if we replace a column by itself minus multiples of other columns. Apply $C_1 \to C_1 - C_2 - 4C_3$:

    $C_1 \text{ becomes } \begin{pmatrix} 4 - 4 \\ 4 - 4 \\ 4 - 4\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$ …

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11. KCET 2018Set A-11 markMCQ
    Q.VERSION: 13-A 32. If $f(x) = \begin{cases} \dfrac{\log_e x}{x-1} & x \neq 1 \\ k & x = 1 \end{cases}$ is continuous at $x = 1$, then the value of $k$ is (A) $e$ (B) $1$ (C) $-1$ (D) $0$

    ›Reveal solutionSolution

    Continuity means $k$ must equal $\lim_{x\to1}\frac{\ln x}{x-1}$, a $\frac00$ form whose value is the standard limit $1$.

    Step 1 — The condition for continuity.

    $f$ is continuous at $x = 1$ iff

    $\lim_{x \to 1} f(x) = f(1) = k$

    So we must evaluate $L = \displaystyle\lim_{x\to1}\frac{\log_e x}{x-1}$ and set $k = L$.

    Step 2 — Recognise the indeterminate form.

    As $x \to 1$: numerator $\log_e 1 = 0$, denominator $1-1 = 0$. It is $\frac{0}{0}$, so the limit is not read off by substitution.

    Step 3 — Evaluate (method 1: substitution to a standard limit).

    Put $x = 1 + h$, so $h \to 0$ as $x \to 1$:

    $L = \lim_{h\to0}\frac{\log_e(1+h)}{h} = 1$

    using the standard limit $\lim_{h\to0}\frac{\ln(1+h)}{h} = 1$ (which follows from the series $\ln(1+h) = h - \frac{h^2}{2} + \cdots$). …

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12. KCET 2025Set A-11 markMCQ
    Q.$\lim_{x \to 1} \dfrac{x^4 - \sqrt{x}}{\sqrt{x} - 1}$ is (A) $0$ (B) $7$ (C) Does not exist (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    Substitute $t=\sqrt x$ to clear the radicals, factor out the common $t$, and use the standard limit $\lim_{t\to1}\frac{t^n-1}{t-1} = n$.

    Step 1 — Recognise the indeterminate form

    At $x = 1$: numerator $= 1^4 - \sqrt1 = 0$, denominator $= \sqrt1 - 1 = 0$. So the limit is of the form $\dfrac{0}{0}$ — it exists (option (C) is a trap) but must be resolved by cancelling the common factor.

    Step 2 — Substitute to remove the radicals

    Let

    $t = \sqrt{x} \quad\Longrightarrow\quad x = t^2, \qquad x \to 1 \Rightarrow t \to 1$

    Then $x^4 = t^8$ and $\sqrt{x} = t$, so

    $L = \lim_{x\to1}\frac{x^4 - \sqrt x}{\sqrt x - 1} = \lim_{t\to1}\frac{t^8 - t}{t - 1}$

    Everything is now a polynomial — much easier to factor.

    Step 3 — Factor and split

    $\frac{t^8 - t}{t-1} = \frac{t\,(t^7 - 1)}{t - 1} = t \cdot \frac{t^7 - 1}{t - 1}$

    Step 4 — Apply the standard limit

    The standard result (from the binomial/derivative definition) is

    $\lim_{t\to a}\frac{t^n - a^n}{t - a} = n\,a^{\,n-1}$

    With $a = 1$, $n = 7$: …

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