Class 12 Maths · Continuity And Differentiability · Long Answer #135

📋 CBSE Stepwise Marking Scheme

• [0.5 marks] Use the product rule on x³v, where v=(cos x)^x and v′=v[ln(cos x)−x tan x]. The inverse-sine square-root term contributes 1/[2√x√(1−x)].
• [1 mark] Applying the indicated rules gives y′=\(x^{3} \left(- \frac{x \sin{\left(x \right)}}{\cos{\left(x \right)}} + \log{\left(\cos{\left(x \right)} \right)}\right) \cos^{x}{\left(x \right)} + 3 x^{2} \cos^{x}{\left(x \right)} + \frac{1}{2 \sqrt{x} \sqrt{1 - x}}\).
• [1 mark] Collecting factors gives y′=\(\frac{- 2 x^{\frac{9}{2}} \sqrt{1 - x} \sin{\left(x \right)} \cos^{x}{\left(x \right)} + 2 x^{\frac{7}{2}} \sqrt{1 - x} \log{\left(\cos{\left(x \right)} \right)} \cos{\left(x \right)} \cos^{x}{\left(x \right)} + 6 x^{\frac{5}{2}} \sqrt{1 - x} \cos{\left(x \right)} \cos^{x}{\left(x \right)} + \cos{\left(x \right)}}{2 \sqrt{x} \sqrt{1 - x} \cos{\left(x \right)}}\).
• [0.5 marks] Therefore dy/dx=\(x^{3} \left(- \frac{x \sin{\left(x \right)}}{\cos{\left(x \right)}} + \log{\left(\cos{\left(x \right)} \right)}\right) \cos^{x}{\left(x \right)} + 3 x^{2} \cos^{x}{\left(x \right)} + \frac{1}{2 \sqrt{x} \sqrt{1 - x}}\).

Master Chapter 5: Continuity And Differentiability