Question
Evaluate for \(\frac{dy}{dx} \,,\, y = \ln(x + \sqrt{1 + x^2 })\)
Solution:
Using chain rules,
\(\frac{dy}{dx} = \frac{1}{x + \sqrt{1 + x^2}}[(1) +( \frac{1}{2} \frac{1}{\sqrt{1 + x^2}} \cdot \{(0) + (2x)\})]\)
\(\Rightarrow \frac{dy}{dx} = \frac{1}{x + \sqrt{1 + x^2}}[\frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}}]\)
\(\Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}}\)