Question

Show that \(\frac{d}{{dx}}\ln \left( {x + \sqrt {{x^2} + 1} } \right) = \frac{1}{{\sqrt {{x^2} + 1} }}\).

Short answer

It is proved that \(\frac{d}{{dx}}\left( {\ln \left( {x + \sqrt {{x^2} + 1} } \right)} \right) = \frac{1}{{\sqrt {{x^2} + 1} }}\).

Step-by-step solution

Derivative of logarithmic functions

The derivative of a logarithmicfunctionis shown below:

1. \(\frac{d}{{dx}}\left( {{{\log }_b}x} \right) = \frac{1}{{x\ln b}}\)
2. \(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)
3. \(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\) or \(\frac{d}{{dx}}\left( {\ln g\left( x \right)} \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}\)

Show that \(\frac{d}{{dx}}\ln \left( {x + \sqrt {{x^2} + 1} } \right) = \frac{1}{{\sqrt {{x^2} + 1} }}\)

Evaluate the derivative of the function as shown below:

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {\ln \left( {x + \sqrt {{x^2} + 1} } \right)} \right)&= \frac{1}{{x + \sqrt {{x^2} + 1} }} \cdot \frac{d}{{dx}}\left( {x + \sqrt {{x^2} + 1} } \right)\\&= \frac{1}{{x + \sqrt {{x^2} + 1} }} \cdot \left( {1 + \frac{1}{{2\sqrt {{x^2} + 1} }} \cdot \frac{d}{{dx}}\left( {{x^2}} \right)} \right)\\&= \frac{1}{{x + \sqrt {{x^2} + 1} }} \cdot \left( {1 + \frac{{2x}}{{2\sqrt {{x^2} + 1} }}} \right)\\&= \frac{1}{{x + \sqrt {{x^2} + 1} }} \cdot \left( {1 + \frac{x}{{\sqrt {{x^2} + 1} }}} \right)\end{aligned}\)

Simplify further,

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {\ln \left( {x + \sqrt {{x^2} + 1} } \right)} \right)&= \frac{1}{{x + \sqrt {{x^2} + 1} }} \cdot \left( {\frac{{x + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + 1} }}} \right)\\&= \frac{1}{{\sqrt {{x^2} + 1} }}\end{aligned}\)

Thus, it is proved that \(\frac{d}{{dx}}\left( {\ln \left( {x + \sqrt {{x^2} + 1} } \right)} \right) = \frac{1}{{\sqrt {{x^2} + 1} }}\).