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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}

Final Answer


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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