Class 12 Maths · Continuity And Differentiability · Long Answer #83

📋 CBSE Stepwise Marking Scheme

• [0.5 marks] Use the chain rule for e^(x²cos x), differentiating its exponent as 2x cos x−x²sin x. For v=(cos x)^x use ln v=x ln(cos x), giving v′=v[ln(cos x)−x tan x].
• [1 mark] Applying the indicated rules gives y′=\(\left(- x^{2} \sin{\left(x \right)} + 2 x \cos{\left(x \right)}\right) e^{x^{2} \cos{\left(x \right)}} + \left(- \frac{x \sin{\left(x \right)}}{\cos{\left(x \right)}} + \log{\left(\cos{\left(x \right)} \right)}\right) \cos^{x}{\left(x \right)}\).
• [1 mark] Collecting factors gives y′=\(- \frac{x^{2} e^{x^{2} \cos{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)} - 2 x e^{x^{2} \cos{\left(x \right)}} \cos^{2}{\left(x \right)} + x \sin{\left(x \right)} \cos^{x}{\left(x \right)} - \log{\left(\cos{\left(x \right)} \right)} \cos{\left(x \right)} \cos^{x}{\left(x \right)}}{\cos{\left(x \right)}}\).
• [0.5 marks] Therefore dy/dx=\(\left(- x^{2} \sin{\left(x \right)} + 2 x \cos{\left(x \right)}\right) e^{x^{2} \cos{\left(x \right)}} + \left(- \frac{x \sin{\left(x \right)}}{\cos{\left(x \right)}} + \log{\left(\cos{\left(x \right)} \right)}\right) \cos^{x}{\left(x \right)}\).

Master Chapter 5: Continuity And Differentiability