Questions and Answers (Board Exam)

Board Exam Strategy

This section contains crucial Previous Year Questions (PYQs) from various board examinations, including recent-style board questions.

To score full marks in board exams, remember:

• State the Condition Clearly: For continuity questions, always write: "For $f(x)$ to be continuous at $x=c$, we must have $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$."
• For Differentiability: First state that differentiability implies continuity. Then write the formal definitions of LHD and RHD before evaluating them.
• Implicit Differentiation: Explicitly write: "Differentiating both sides with respect to $x$" and show the product rule / chain rule wherever $y$ is involved.
• Logarithmic Differentiation: For expressions like $y = u+v$, clearly mention that logarithm of a sum cannot be taken directly, i.e. $\ln(u+v) \neq \ln u + \ln v$.
• Second Derivative Proofs: When proving equations involving $y_1$ and $y_2$, first find $y_1$, then differentiate again carefully using the product rule and chain rule.

Question 1 [CBSE 2026]

Find the value of $k$ so that the function $f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases}$ is continuous at $x = 2$.

Solution: Step 1: For continuity at $x=2$, we need $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2).$$

Step 2: Find the left hand limit. For $x \le 2$, $f(x)=kx^2$. Hence, $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} kx^2 = 4k.$$

Step 3: Find the right hand limit. For $x>2$, $f(x)=3$. Hence, $$\lim_{x \to 2^+} f(x)=3.$$

Step 4: Find the actual value of the function at $x=2$. Since the branch for $x \le 2$ includes $x=2$, $$f(2)=k(2)^2=4k.$$

Step 5: Equate the three values for continuity. $$4k=3.$$ Therefore, $$k=\frac{3}{4}.$$

Answer: $k = \frac{3}{4}$

Question 2 [CBSE 2024]

Find the values of $a$ and $b$ such that the function defined by $f(x) = \begin{cases} 5, & \text{if } x \le 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \ge 10 \end{cases}$ is a continuous function.

Solution: Step 1: The function can fail to be continuous only at the break points $x=2$ and $x=10$. So we check continuity at these two points.

Step 2: Continuity at $x=2$. Left hand limit: $$\lim_{x \to 2^-} f(x)=5.$$ Right hand limit: $$\lim_{x \to 2^+} f(x)=2a+b.$$ Also, $$f(2)=5.$$ Hence, $$2a+b=5. \qquad ...(1)$$

Step 3: Continuity at $x=10$. Left hand limit: $$\lim_{x \to 10^-} f(x)=10a+b.$$ Right hand limit: $$\lim_{x \to 10^+} f(x)=21.$$ Also, $$f(10)=21.$$ Hence, $$10a+b=21. \qquad ...(2)$$

Step 4: Solve equations (1) and (2). Subtract (1) from (2): $$(10a+b)-(2a+b)=21-5$$ $$8a=16$$ $$a=2.$$ Substitute in (1): $$2(2)+b=5$$ $$4+b=5$$ $$b=1.$$

Answer: $a = 2,\; b = 1$

Question 3 [CBSE 2025]

Find the value of $k$ so that the function $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$.

Solution: Step 1: For continuity at $x=\frac{\pi}{2}$, we need $$\lim_{x \to \pi/2} \frac{k\cos x}{\pi-2x}=f\left(\frac{\pi}{2}\right)=3.$$

Step 2: Evaluate the limit. Let $$x=\frac{\pi}{2}-h,$$ so that as $x \to \frac{\pi}{2}$, we have $h \to 0$. Then, $$\cos x = \cos\left(\frac{\pi}{2}-h\right)=\sin h,$$ and $$\pi-2x = \pi - 2\left(\frac{\pi}{2}-h\right)=2h.$$ Therefore, $$\lim_{x \to \pi/2} \frac{k\cos x}{\pi-2x} = \lim_{h \to 0} \frac{k\sin h}{2h} = \frac{k}{2}\lim_{h \to 0} \frac{\sin h}{h}.$$ Using $$\lim_{h \to 0} \frac{\sin h}{h}=1,$$ we get $$\lim_{x \to \pi/2} \frac{k\cos x}{\pi-2x}=\frac{k}{2}.$$

Step 3: Equate with the function value. $$\frac{k}{2}=3$$ $$k=6.$$

Answer: $k = 6$

Question 4 [CBSE 2022]

Discuss the continuity of the function $f(x) = |x - 1| + |x - 2|$ at $x = 1$ and $x = 2$.

Solution: Step 1: Remove modulus signs by considering intervals.

For $x<1$: $$|x-1|=-(x-1), \qquad |x-2|=-(x-2)$$ So, $$f(x)=-(x-1)-(x-2)=-2x+3.$$

For $1 \le x < 2$: $$|x-1|=x-1, \qquad |x-2|=-(x-2)$$ So, $$f(x)=(x-1)-(x-2)=1.$$

For $x \ge 2$: $$|x-1|=x-1, \qquad |x-2|=x-2$$ So, $$f(x)=x-1+x-2=2x-3.$$

Step 2: Check continuity at $x=1$. Left hand limit: $$\lim_{x \to 1^-} f(x)=\lim_{x \to 1^-} (-2x+3)=1.$$ Right hand limit: $$\lim_{x \to 1^+} f(x)=\lim_{x \to 1^+} 1=1.$$ Function value: $$f(1)=|1-1|+|1-2|=0+1=1.$$ Hence, $$\lim_{x \to 1^-} f(x)=\lim_{x \to 1^+} f(x)=f(1),$$ so the function is continuous at $x=1$.

Step 3: Check continuity at $x=2$. Left hand limit: $$\lim_{x \to 2^-} f(x)=1.$$ Right hand limit: $$\lim_{x \to 2^+} f(x)=\lim_{x \to 2^+}(2x-3)=1.$$ Function value: $$f(2)=|2-1|+|2-2|=1+0=1.$$ Therefore, the function is continuous at $x=2$ also.

Answer: $f(x)$ is continuous at both $x = 1$ and $x = 2$.

Question 5 [CBSE 2026]

Determine the value of $k$ for which the function $f(x) = \begin{cases} \frac{1 - \cos 2x}{2x^2}, & \text{if } x \ne 0 \\ k, & \text{if } x = 0 \end{cases}$ is continuous at $x = 0$.

Solution: Step 1: For continuity at $x=0$, we need $$\lim_{x \to 0} \frac{1-\cos 2x}{2x^2}=k.$$

Step 2: Use the identity $$1-\cos 2x = 2\sin^2 x.$$ So, $$\lim_{x \to 0} \frac{1-\cos 2x}{2x^2} = \lim_{x \to 0} \frac{2\sin^2 x}{2x^2} = \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2.$$

Step 3: Use the standard limit $$\lim_{x \to 0} \frac{\sin x}{x}=1.$$ Hence, $$\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = 1.$$ Thus, for continuity, $$k=1.$$

Answer: $k = 1$

Question 6 [CBSE 2023]

Show that the function $f(x) = x - [x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer function.

Solution: Step 1: Let $c$ be any integer. We examine the function at $x=c$.

Step 2: Find the left hand limit. As $x \to c^-$, $x$ is slightly less than $c$, so $$[x]=c-1.$$ Hence, $$\lim_{x \to c^-} (x-[x]) = \lim_{x \to c^-} (x-(c-1)) = c-(c-1)=1.$$

Step 3: Find the right hand limit. As $x \to c^+$, $x$ is slightly greater than $c$, so $$[x]=c.$$ Hence, $$\lim_{x \to c^+} (x-[x]) = \lim_{x \to c^+} (x-c)=c-c=0.$$

Step 4: Compare the limits. Since $$\lim_{x \to c^-} f(x)=1 \neq 0=\lim_{x \to c^+} f(x),$$ the left hand limit and right hand limit are unequal. Therefore, the function is discontinuous at every integral point.

Answer: Proved analytically.

Question 7 [CBSE 2024]

Show that the function $f(x) = |x - 3|$ is continuous but not differentiable at $x = 3$.

Solution: Step 1: Check continuity at $x=3$. Left hand limit: For $x<3$, $|x-3|=-(x-3)$, so $$\lim_{x \to 3^-} |x-3| = \lim_{x \to 3^-} (-(x-3))=0.$$ Right hand limit: For $x>3$, $|x-3|=x-3$, so $$\lim_{x \to 3^+} |x-3| = \lim_{x \to 3^+}(x-3)=0.$$ Function value: $$f(3)=|3-3|=0.$$ Hence, $$\lim_{x \to 3^-} f(x)=\lim_{x \to 3^+} f(x)=f(3)=0,$$ so the function is continuous at $x=3$.

Step 2: Check differentiability using LHD and RHD. For $h<0$, $f(3+h)=|h|=-h$, so $$LHD = \lim_{h \to 0^-} \frac{f(3+h)-f(3)}{h} = \lim_{h \to 0^-} \frac{-h-0}{h}=-1.$$ For $h>0$, $f(3+h)=|h|=h$, so $$RHD = \lim_{h \to 0^+} \frac{f(3+h)-f(3)}{h} = \lim_{h \to 0^+} \frac{h-0}{h}=1.$$

Step 3: Since $$LHD \ne RHD,$$ $f(x)$ is not differentiable at $x=3$.

Answer: Proved.

Question 8 [CBSE 2026]

Find the values of $p$ and $q$ if $f(x) = \begin{cases} x^2 + 3x + p, & \text{if } x \le 1 \\ qx + 2, & \text{if } x > 1 \end{cases}$ is differentiable at $x = 1$.

Solution: Step 1: If a function is differentiable at a point, then it must first be continuous there. So, continuity at $x=1$ gives $$\lim_{x \to 1^-} f(x)=\lim_{x \to 1^+} f(x)=f(1).$$ From the left branch, $$\lim_{x \to 1^-} f(x)=1^2+3(1)+p=4+p.$$ From the right branch, $$\lim_{x \to 1^+} f(x)=q(1)+2=q+2.$$ Thus, $$4+p=q+2. \qquad ...(1)$$

Step 2: For differentiability, LHD = RHD. Derivative of the left branch: $$\frac{d}{dx}(x^2+3x+p)=2x+3.$$ So at $x=1$, $$LHD=2(1)+3=5.$$ Derivative of the right branch: $$\frac{d}{dx}(qx+2)=q.$$ So, $$RHD=q.$$ Hence, $$q=5.$$

Step 3: Put $q=5$ in equation (1): $$4+p=5+2$$ $$4+p=7$$ $$p=3.$$

Answer: $p = 3,\; q = 5$

Question 9 [CBSE 2025]

Check whether the function $f(x) = x|x|$ is differentiable at $x = 0$.

Solution: Step 1: Rewrite the function in piecewise form. For $x<0$, $|x|=-x$, so $$f(x)=x(-x)=-x^2.$$ For $x \ge 0$, $|x|=x$, so $$f(x)=x(x)=x^2.$$ Thus, $$f(x)=\begin{cases} -x^2, & x<0 \\ x^2, & x \ge 0 \end{cases}$$

Step 2: Find the left hand derivative at $x=0$. $$LHD = \lim_{h \to 0^-} \frac{f(h)-f(0)}{h}.$$ Since $f(0)=0$ and for $h<0$, $f(h)=-h^2$, we get $$LHD = \lim_{h \to 0^-} \frac{-h^2}{h} = \lim_{h \to 0^-} (-h)=0.$$

Step 3: Find the right hand derivative. For $h>0$, $f(h)=h^2$, so $$RHD = \lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} h=0.$$

Step 4: Since $$LHD=RHD=0,$$ the function is differentiable at $x=0$. Also, $$f'(0)=0.$$

Answer: Yes, it is differentiable at $x = 0$.

Question 10 [CBSE 2023]

Differentiate $y = \cos(\sqrt{x})$ with respect to $x$.

Solution: Step 1: The outer function is $\cos u$ and the inner function is $u=\sqrt{x}$. By chain rule, $$\frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}).$$

Step 2: Differentiate $\sqrt{x}=x^{1/2}$. $$\frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}.$$

Step 3: Substitute back. $$\frac{dy}{dx} = -\sin(\sqrt{x})\cdot \frac{1}{2\sqrt{x}} = -\frac{\sin(\sqrt{x})}{2\sqrt{x}}.$$

Answer: $\frac{-\sin(\sqrt{x})}{2\sqrt{x}}$

Question 11 [CBSE 2026]

Differentiate $\tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right)$ with respect to $x$.

Solution: Step 1: Use the identity $$\frac{\sin x}{1+\cos x}=\tan\frac{x}{2},$$ where the expression is defined. Thus, $$y=\tan^{-1}\left(\tan\frac{x}{2}\right).$$

Step 2: For the standard principal-value branch used in such board problems, this simplifies to $$y=\frac{x}{2}.$$

Step 3: Differentiate with respect to $x$: $$\frac{dy}{dx}=\frac{1}{2}.$$

Answer: $\frac{1}{2}$

Question 12 [CBSE 2024]

Find $\frac{dy}{dx}$ if $y = \sin^{-1} \left( \frac{2^{x+1}}{1 + 4^x} \right)$.

Solution: Step 1: Rewrite the inside expression. Since $$2^{x+1}=2\cdot 2^x, \qquad 4^x=(2^x)^2,$$ we have $$y=\sin^{-1}\left(\frac{2\cdot 2^x}{1+(2^x)^2}\right).$$

Step 2: Compare with the identity $$\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}.$$ Let $$2^x=\tan\theta.$$ Then $$\theta=\tan^{-1}(2^x).$$ Therefore, $$y=\sin^{-1}(\sin 2\theta)=2\theta=2\tan^{-1}(2^x)$$ for the principal branch used in the problem.

Step 3: Differentiate. $$\frac{dy}{dx} = 2\cdot \frac{1}{1+(2^x)^2}\cdot \frac{d}{dx}(2^x).$$ Now, $$\frac{d}{dx}(2^x)=2^x\ln 2.$$ Hence, $$\frac{dy}{dx}=2\cdot \frac{2^x\ln 2}{1+4^x} = \frac{2^{x+1}\ln 2}{1+4^x}.$$

Answer: $\frac{2^{x+1} \ln 2}{1 + 4^x}$

Question 13 [CBSE 2022]

Find $\frac{dy}{dx}$ if $y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right)$ where $0 < x < \frac{1}{\sqrt{2}}$.

Solution: Step 1: The expression $2x^2-1$ suggests the identity $$2\cos^2\theta-1=\cos 2\theta.$$ Let $$x=\cos\theta.$$ Then $$\theta=\cos^{-1}x.$$

Step 2: Substitute into the function: $$y=\sec^{-1}\left(\frac{1}{2\cos^2\theta-1}\right)=\sec^{-1}\left(\frac{1}{\cos 2\theta}\right)=\sec^{-1}(\sec 2\theta).$$ In the given domain, this gives $$y=2\theta=2\cos^{-1}x.$$

Step 3: Differentiate: $$\frac{dy}{dx}=2\cdot \frac{d}{dx}(\cos^{-1}x)=2\left(-\frac{1}{\sqrt{1-x^2}}\right)=-\frac{2}{\sqrt{1-x^2}}.$$

Answer: $\frac{-2}{\sqrt{1 - x^2}}$

Question 14 [CBSE 2025]

Find the derivative of $\tan^{-1} \left( \frac{\sqrt{1+x^2} - 1}{x} \right)$ with respect to $x$.

Solution: Step 1: Let $$x=\tan\theta.$$ Then $$\theta=\tan^{-1}x$$ and $$\sqrt{1+x^2}=\sqrt{1+\tan^2\theta}=\sec\theta.$$ Thus, $$y=\tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right).$$

Step 2: Simplify the inner expression: $$\frac{\sec\theta-1}{\tan\theta} = \frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}} = \frac{1-\cos\theta}{\sin\theta}.$$ Now use half-angle identities: $$1-\cos\theta=2\sin^2\frac{\theta}{2}, \qquad \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.$$ So, $$\frac{1-\cos\theta}{\sin\theta}=\tan\frac{\theta}{2}.$$ Hence, $$y=\tan^{-1}\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x.$$

Step 3: Differentiate with respect to $x$: $$\frac{dy}{dx}=\frac{1}{2}\cdot \frac{1}{1+x^2} = \frac{1}{2(1+x^2)}.$$

Answer: $\frac{1}{2(1+x^2)}$

Question 15 [CBSE 2026]

Find $\frac{dy}{dx}$ if $x^2 + xy + y^2 = 100$.

Solution: Step 1: Differentiate both sides with respect to $x$: $$\frac{d}{dx}(x^2)+\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=0.$$

Step 2: Apply product rule and chain rule: $$2x + \left(x\frac{dy}{dx}+y\right) + 2y\frac{dy}{dx}=0.$$

Step 3: Collect the terms containing $\frac{dy}{dx}$: $$\left(x+2y\right)\frac{dy}{dx}=-(2x+y).$$

Step 4: Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{-(2x+y)}{x+2y}.$$

Answer: $\frac{-(2x + y)}{x + 2y}$

Question 16 [CBSE 2023]

Find $\frac{dy}{dx}$ if $\sin^2 y + \cos(xy) = k$.

Solution: Step 1: Differentiate both sides with respect to $x$. Since $k$ is a constant, $$\frac{d}{dx}(\sin^2 y)+\frac{d}{dx}(\cos(xy))=0.$$

Step 2: Differentiate each term. For $\sin^2 y$: $$\frac{d}{dx}(\sin^2 y)=2\sin y\cos y\frac{dy}{dx}=\sin 2y\frac{dy}{dx}.$$ For $\cos(xy)$: $$\frac{d}{dx}(\cos(xy))=-\sin(xy)\cdot \frac{d}{dx}(xy).$$ Using product rule, $$\frac{d}{dx}(xy)=x\frac{dy}{dx}+y.$$ So, $$\frac{d}{dx}(\cos(xy))=-\sin(xy)\left(x\frac{dy}{dx}+y\right).$$

Step 3: Substitute and simplify: $$\sin 2y\frac{dy}{dx} - x\sin(xy)\frac{dy}{dx} - y\sin(xy)=0.$$

Step 4: Factor out $\frac{dy}{dx}$: $$\frac{dy}{dx}\left(\sin 2y - x\sin(xy)\right)=y\sin(xy).$$ Hence, $$\frac{dy}{dx}=\frac{y\sin(xy)}{\sin 2y - x\sin(xy)}.$$

Answer: $\frac{y\sin(xy)}{\sin 2y - x\sin(xy)}$

Question 17 [CBSE 2024]

If $y = \sqrt{\sin x + y}$, find $\frac{dy}{dx}$.

Solution: Step 1: Square both sides to remove the square root: $$y^2 = \sin x + y.$$

Step 2: Differentiate implicitly with respect to $x$: $$2y\frac{dy}{dx}=\cos x + \frac{dy}{dx}.$$

Step 3: Bring the derivative terms together: $$2y\frac{dy}{dx} - \frac{dy}{dx}=\cos x.$$ $$(2y-1)\frac{dy}{dx}=\cos x.$$

Step 4: Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{\cos x}{2y-1}.$$

Answer: $\frac{\cos x}{2y - 1}$

Question 18 [CBSE 2025]

If $x \sqrt{1 + y} + y \sqrt{1 + x} = 0$, for $-1 < x < 1, x \ne y$, prove that $\frac{dy}{dx} = \frac{-1}{(1+x)^2}$.

Solution: Step 1: Rearrange the given relation: $$x\sqrt{1+y} = -y\sqrt{1+x}.$$

Step 2: Square both sides: $$x^2(1+y)=y^2(1+x).$$ Expanding, $$x^2+x^2y = y^2+xy^2.$$

Step 3: Bring all terms to one side: $$x^2-y^2+x^2y-xy^2=0.$$ Factor: $$(x-y)(x+y)+xy(x-y)=0.$$ $$(x-y)(x+y+xy)=0.$$ Since $x \ne y$, we must have $$x+y+xy=0.$$

Step 4: Solve explicitly for $y$: $$y(1+x)=-x$$ $$y=-\frac{x}{1+x}.$$

Step 5: Differentiate using the quotient rule: $$\frac{dy}{dx} = \frac{(1+x)(-1)-(-x)(1)}{(1+x)^2} = \frac{-1-x+x}{(1+x)^2} = -\frac{1}{(1+x)^2}.$$

Answer: Proved.

Question 19 [CBSE 2026]

Find $\frac{dy}{dx}$ if $y = x^x - 2^{\sin x}$.

Solution: Step 1: Since the expression is a difference, logarithmic differentiation cannot be applied directly to the whole function. Write $$y=u-v,$$ where $$u=x^x, \qquad v=2^{\sin x}.$$ Then, $$\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}.$$

Step 2: Differentiate $u=x^x$ by logarithmic differentiation. Take logarithm: $$\ln u = x\ln x.$$ Differentiate: $$\frac{1}{u}\frac{du}{dx}=1+\ln x.$$ Thus, $$\frac{du}{dx}=u(1+\ln x)=x^x(1+\ln x).$$

Step 3: Differentiate $v=2^{\sin x}$. Using $$\frac{d}{dx}(a^{f(x)})=a^{f(x)}\ln a\cdot f'(x),$$ we get $$\frac{dv}{dx}=2^{\sin x}\ln 2\cdot \cos x.$$

Step 4: Combine the two derivatives: $$\frac{dy}{dx}=x^x(1+\ln x)-2^{\sin x}\cos x\ln 2.$$

Answer: $x^x(1 + \ln x) - 2^{\sin x} \cos x \ln 2$

Question 20 [CBSE 2022]

Find $\frac{dy}{dx}$ of the function $x^y = y^x$.

Solution: Step 1: Take natural logarithm on both sides: $$\ln(x^y)=\ln(y^x).$$ So, $$y\ln x = x\ln y.$$

Step 2: Differentiate both sides with respect to $x$. Using product rule, $$\frac{dy}{dx}\ln x + y\cdot \frac{1}{x} = 1\cdot \ln y + x\cdot \frac{1}{y}\frac{dy}{dx}.$$ That is, $$\ln x\frac{dy}{dx} + \frac{y}{x} = \ln y + \frac{x}{y}\frac{dy}{dx}.$$

Step 3: Collect the terms containing $\frac{dy}{dx}$ on one side: $$\left(\ln x - \frac{x}{y}\right)\frac{dy}{dx} = \ln y - \frac{y}{x}.$$

Step 4: Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{\ln y - \frac{y}{x}}{\ln x - \frac{x}{y}}.$$ Multiplying numerator and denominator by $xy$, $$\frac{dy}{dx} = \frac{y(x\ln y - y)}{x(y\ln x - x)}.$$

Answer: $\frac{y(x\ln y - y)}{x(y\ln x - x)}$

Question 21 [CBSE 2024]

Differentiate $(\sin x)^x + \sin^{-1}\sqrt{x}$ with respect to $x$.

Solution: Step 1: Let $$y=u+v,$$ where $$u=(\sin x)^x, \qquad v=\sin^{-1}(\sqrt{x}).$$ Then, $$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}.$$

Step 2: Differentiate $u=(\sin x)^x$ by logarithmic differentiation. Take logarithm: $$\ln u = x\ln(\sin x).$$ Differentiate: $$\frac{1}{u}\frac{du}{dx} = \ln(\sin x) + x\cdot \frac{1}{\sin x}\cos x.$$ So, $$\frac{du}{dx} = (\sin x)^x\left(\ln(\sin x)+x\cot x\right).$$

Step 3: Differentiate $v=\sin^{-1}(\sqrt{x})$. Using chain rule, $$\frac{dv}{dx} = \frac{1}{\sqrt{1-(\sqrt{x})^2}}\cdot \frac{d}{dx}(\sqrt{x}) = \frac{1}{\sqrt{1-x}}\cdot \frac{1}{2\sqrt{x}}.$$ Thus, $$\frac{dv}{dx}=\frac{1}{2\sqrt{x}\sqrt{1-x}} = \frac{1}{2\sqrt{x-x^2}}.$$

Step 4: Add the two derivatives: $$\frac{dy}{dx} = (\sin x)^x\left(x\cot x + \ln(\sin x)\right) + \frac{1}{2\sqrt{x-x^2}}.$$

Answer: $(\sin x)^x (x\cot x + \ln(\sin x)) + \frac{1}{2\sqrt{x-x^2}}$

Question 22 [CBSE 2025]

Find the derivative of $x^{\sin x} + (\sin x)^{\cos x}$ w.r.t. $x$.

Solution: Step 1: Write $$y=u+v,$$ where $$u=x^{\sin x}, \qquad v=(\sin x)^{\cos x}.$$ Then, $$\frac{dy}{dx}=u'+v'.$$

Step 2: Differentiate $u=x^{\sin x}$. Take logarithm: $$\ln u = \sin x\ln x.$$ Differentiate: $$\frac{u'}{u}=\cos x\ln x + \sin x\cdot \frac{1}{x}.$$ So, $$u' = x^{\sin x}\left(\cos x\ln x + \frac{\sin x}{x}\right).$$

Step 3: Differentiate $v=(\sin x)^{\cos x}$. Take logarithm: $$\ln v = \cos x\ln(\sin x).$$ Differentiate: $$\frac{v'}{v} = -\sin x\ln(\sin x) + \cos x\cdot \frac{1}{\sin x}\cos x.$$ That is, $$\frac{v'}{v} = -\sin x\ln(\sin x) + \cos x\cot x.$$ Hence, $$v' = (\sin x)^{\cos x}\left(\cos x\cot x - \sin x\ln(\sin x)\right).$$

Step 4: Therefore, $$\frac{dy}{dx} = x^{\sin x}\left(\frac{\sin x}{x}+\cos x\ln x\right) + (\sin x)^{\cos x}\left(\cos x\cot x-\sin x\ln(\sin x)\right).$$

Answer: $x^{\sin x} \left(\frac{\sin x}{x} + \cos x \ln x\right) + (\sin x)^{\cos x} [\cos x \cot x - \sin x \ln(\sin x)]$

Question 23 [CBSE 2026]

Find $\frac{dy}{dx}$ if $y = (\log x)^x + x^{\log x}$.

Solution: Step 1: Let $$y=u+v,$$ where $$u=(\log x)^x, \qquad v=x^{\log x}.$$ Here $\log x$ is taken as natural logarithm. Then, $$\frac{dy}{dx}=u'+v'.$$

Step 2: Differentiate $u=(\log x)^x$. Take logarithm: $$\ln u = x\ln(\log x).$$ Differentiate: $$\frac{u'}{u} = \ln(\log x) + x\cdot \frac{1}{\log x}\cdot \frac{1}{x} = \ln(\log x)+\frac{1}{\log x}.$$ So, $$u'=(\log x)^x\left(\ln(\log x)+\frac{1}{\log x}\right).$$

Step 3: Differentiate $v=x^{\log x}$. Take logarithm: $$\ln v = (\log x)(\ln x)=(\ln x)^2.$$ Differentiate: $$\frac{v'}{v}=2\ln x\cdot \frac{1}{x}.$$ Thus, $$v'=x^{\log x}\cdot \frac{2\ln x}{x}.$$

Step 4: Add the two derivatives: $$\frac{dy}{dx} = (\log x)^x\left[\frac{1}{\log x}+\ln(\log x)\right] + x^{\log x}\left(\frac{2\ln x}{x}\right).$$

Answer: $(\log x)^x \left[ \frac{1}{\log x} + \ln(\log x) \right] + x^{\log x} \left(\frac{2\ln x}{x}\right)$