CBSE Class 12 Mathematics 2025: question 28

Hints

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Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Let $u=\frac{x^3}{3}-1$. By the chain rule, the outer square root gives $\frac{dy}{dx}=\frac{1}{2\sqrt{\log \{\sin u\}}}\cdot \frac{d}{dx}\log (\sin u)$.

   1 mark

2. $\frac{d}{dx}\log (\sin u)=\frac{1}{\sin u}\cdot \cos u\cdot \frac{du}{dx}$

   1 mark

3. $\frac{du}{dx}=\frac{3x^2}{3}=x^2$

   ½ mark

4. $\frac{dy}{dx}=\frac{x^2\cot \left(\frac{x^3}{3}-1\right)}{2\sqrt{\log \left\{\sin \left(\frac{x^3}{3}-1\right)\right\}}}$

   ½ mark

Answer: $\frac{dy}{dx}=\frac{x^2\cot \left(\frac{x^3}{3}-1\right)}{2\sqrt{\log \{\sin (\frac{x^3}{3}-1)\}}}$

Revise the chapter

• Chapter: Continuity and Differentiability (notes, formulas, where marks are lost and every past question)