# Calculating Derivative – A fraction with square root in ln – Exercise 6357

Exercise

Find the derivative of the following function:

$$f(x)=\ln\frac{\sqrt{x^2+1}-1}{x}$$

$$f'(x)=\frac{1}{x\sqrt{x^2+1}}$$

Solution

$$f(x)=\ln\frac{\sqrt{x^2+1}-1}{x}$$

Using Derivative formulas and the quotient rule and chain rule in Derivative Rules, we get the derivative:

$$f'(x)=\frac{1}{\frac{\sqrt{x^2+1}-1}{x}}\cdot \frac{\frac{1}{2\sqrt{x^2+1}}\cdot 2x\cdot x-(\sqrt{x^2+1}-1)\cdot 1}{x^2}=$$

One can simplify the derivative:

$$=\frac{x}{\sqrt{x^2+1}-1}\cdot\frac{1}{\sqrt{x^2+1}}-\frac{\sqrt{x^2+1}-1}{x^2}=$$

$$=\frac{x}{\sqrt{x^2+1}-1}\cdot\frac{x^2-x^2-1+\sqrt{x^2+1}}{x^2\sqrt{x^2+1}}=$$

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

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

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