NCERT Class 12 Maths · Chapter 5 Continuity And Differentiability · Exercise 5.5 Question 6

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