Exercise
Find the derivative of the following function:
f ( x ) = ln ( x + x 2 + 1 ) f(x)=\ln (x+\sqrt{x^2+1}) f ( x ) = ln ( x + x 2 + 1 )
Final Answer
Show final answer
f'(x)=\frac{1}{\sqrt{x^2+1}}
Solution
f ( x ) = ln ( x + x 2 + 1 ) f(x)=\ln (x+\sqrt{x^2+1}) f ( x ) = ln ( x + x 2 + 1 )
Using Derivative formulas Derivative Rules
f ′ ( x ) = 1 x + x 2 + 1 ⋅ ( 1 + 1 2 x 2 + 1 ⋅ 2 x ) = f'(x)=\frac{1}{x+\sqrt{x^2+1}}\cdot (1+\frac{1}{2\sqrt{x^2+1}}\cdot 2x)= f ′ ( x ) = x + x 2 + 1 1 ⋅ ( 1 + 2 x 2 + 1 1 ⋅ 2 x ) =
One can simplify the derivative:
= 1 x + x 2 + 1 ⋅ x 2 + 1 + x x 2 + 1 = =\frac{1}{x+\sqrt{x^2+1}}\cdot\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}= = x + x 2 + 1 1 ⋅ x 2 + 1 x 2 + 1 + x =
= 1 x 2 + 1 =\frac{1}{\sqrt{x^2+1}} = x 2 + 1 1
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