Class 12 Maths · Continuity And Differentiability · Long Answer #136

📋 CBSE Stepwise Marking Scheme

• [0.5 marks] For u=(sin2x)^x, ln u=x ln(sin2x), so u′=u[ln(sin2x)+2x cot2x]. For v=arcsin√(3x), apply the two chain factors 1/√(1−3x) and 3/(2√(3x)).
• [1 mark] Applying the indicated rules gives y′=\(\left(\frac{2 x \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} + \log{\left(\sin{\left(2 x \right)} \right)}\right) \sin^{x}{\left(2 x \right)} + \frac{\sqrt{3}}{2 \sqrt{x} \sqrt{1 - 3 x}}\).
• [1 mark] Collecting factors gives y′=\(\frac{4 x^{\frac{3}{2}} \sqrt{1 - 3 x} \sin^{x}{\left(2 x \right)} \cos{\left(2 x \right)} + 2 \sqrt{x} \sqrt{1 - 3 x} \log{\left(\sin{\left(2 x \right)} \right)} \sin{\left(2 x \right)} \sin^{x}{\left(2 x \right)} + \sqrt{3} \sin{\left(2 x \right)}}{2 \sqrt{x} \sqrt{1 - 3 x} \sin{\left(2 x \right)}}\).
• [0.5 marks] Therefore dy/dx=\(\left(\frac{2 x \cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} + \log{\left(\sin{\left(2 x \right)} \right)}\right) \sin^{x}{\left(2 x \right)} + \frac{\sqrt{3}}{2 \sqrt{x} \sqrt{1 - 3 x}}\).

Master Chapter 5: Continuity And Differentiability