NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.2 Question 7

✅ Step-by-Step Solution

1. Work where cot(x²)>0 and sin(x²)≠0.
2. Power and chain rules: \(\frac{d}{d x} \sqrt{\cot{\left(x^{2} \right)}}\) = \(\frac{1}{2 \sqrt{\cot{\left(x^{2} \right)}}}\) × \(2 x \left(- \cot^{2}{\left(x^{2} \right)} - 1\right)\) = \(\frac{x \left(- \cot^{2}{\left(x^{2} \right)} - 1\right)}{\sqrt{\cot{\left(x^{2} \right)}}}\).
3. Carry the constant factor: \(\frac{d}{d x} 2 \sqrt{\cot{\left(x^{2} \right)}}\) = \(\frac{2 x \left(- \cot^{2}{\left(x^{2} \right)} - 1\right)}{\sqrt{\cot{\left(x^{2} \right)}}}\).
4. Combining the terms gives y′ = \(- \frac{2 x}{\sin^{2}{\left(x^{2} \right)} \sqrt{\cot{\left(x^{2} \right)}}}\).