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Question

# Evaluate for $$\frac{dy}{dx} \,,\, y = \ln(x + \sqrt{1 + x^2 })$$

Solution:

Using chain rules,

$$\frac{dy}{dx} = \frac{1}{x + \sqrt{1 + x^2}}[(1) +( \frac{1}{2} \frac{1}{\sqrt{1 + x^2}} \cdot \{(0) + (2x)\})]$$

$$\Rightarrow \frac{dy}{dx} = \frac{1}{x + \sqrt{1 + x^2}}[\frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}}]$$

$$\Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}}$$