# Calculating Derivative – Rational function in square root inside ln function – Exercise 6932

Exercise

Find the derivative of the function

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

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

Solution

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

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

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

One can simplify the derivative:

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

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

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

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

$$=\frac{1}{x-1}\cdot\frac{1}{x+1}=$$

$$=\frac{1}{(x-1)(x+1)}=$$

Using a short multiplication formula, we get

$$=\frac{1}{x^2-1}$$

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