NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.5 Question 4

✅ Step-by-Step Solution

1. Work for x>0.
2. For a positive base u, write uᵛ=exp(v log u). Its derivative is uᵛ(v′ log u+vu′/u). Here \(\frac{d}{d x} x^{x}\) = \(x^{x} \left(\log{\left(x \right)} + 1\right)\).
3. For a positive base u, write uᵛ=exp(v log u). Its derivative is uᵛ(v′ log u+vu′/u). Here \(\frac{d}{d x} 2^{\sin{\left(x \right)}}\) = \(2^{\sin{\left(x \right)}} \log{\left(2 \right)} \cos{\left(x \right)}\).
4. Carry the constant factor: \(\frac{d}{d x} \left(- 2^{\sin{\left(x \right)}}\right)\) = \(- 2^{\sin{\left(x \right)}} \log{\left(2 \right)} \cos{\left(x \right)}\).
5. Sum rule: \(\frac{d}{d x} \left(- 2^{\sin{\left(x \right)}} + x^{x}\right)\) = \(- 2^{\sin{\left(x \right)}} \log{\left(2 \right)} \cos{\left(x \right)} + x^{x} \left(\log{\left(x \right)} + 1\right)\).
6. Combining the terms gives y′ = \(- 2^{\sin{\left(x \right)}} \log{\left(2 \right)} \cos{\left(x \right)} + x^{x} \left(\log{\left(x \right)} + 1\right)\).