Exercise
Find the derivative of the following function:
f ( x ) = ln x 2 + 1 − 1 x f(x)=\ln\frac{\sqrt{x^2+1}-1}{x} f ( x ) = ln x x 2 + 1 − 1
Final Answer
Show final answer
f'(x)=\frac{1}{x\sqrt{x^2+1}}
Solution
f ( x ) = ln x 2 + 1 − 1 x f(x)=\ln\frac{\sqrt{x^2+1}-1}{x} f ( x ) = ln x x 2 + 1 − 1
Using Derivative formulas Derivative Rules
f ′ ( x ) = 1 x 2 + 1 − 1 x ⋅ 1 2 x 2 + 1 ⋅ 2 x ⋅ x − ( x 2 + 1 − 1 ) ⋅ 1 x 2 = 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}= f ′ ( x ) = x x 2 + 1 − 1 1 ⋅ x 2 2 x 2 + 1 1 ⋅ 2 x ⋅ x − ( x 2 + 1 − 1 ) ⋅ 1 =
One can simplify the derivative:
= x x 2 + 1 − 1 ⋅ 1 x 2 + 1 − x 2 + 1 − 1 x 2 = =\frac{x}{\sqrt{x^2+1}-1}\cdot\frac{1}{\sqrt{x^2+1}}-\frac{\sqrt{x^2+1}-1}{x^2}= = x 2 + 1 − 1 x ⋅ x 2 + 1 1 − x 2 x 2 + 1 − 1 =
= x x 2 + 1 − 1 ⋅ x 2 − x 2 − 1 + x 2 + 1 x 2 x 2 + 1 = =\frac{x}{\sqrt{x^2+1}-1}\cdot\frac{x^2-x^2-1+\sqrt{x^2+1}}{x^2\sqrt{x^2+1}}= = x 2 + 1 − 1 x ⋅ x 2 x 2 + 1 x 2 − x 2 − 1 + x 2 + 1 =
= x x 2 x 2 + 1 = =\frac{x}{x^2\sqrt{x^2+1}}= = x 2 x 2 + 1 x =
= 1 x x 2 + 1 =\frac{1}{x\sqrt{x^2+1}} = x x 2 + 1 1
Have a question? Found a mistake? – Write a comment below!here
