If $f(x) = \left(x + \sqrt{x^{2} + 1}\right)^{5}$, what is $f'\left(\frac{3}{4}\right)$?

Explanation

1. 01

   $f'(x) = 5\left(x + \sqrt{x^{2} + 1}\right)^{4}\left(1 + \frac{x}{\sqrt{x^{2} + 1}}\right) = \frac{5\left(x + \sqrt{x^{2} + 1}\right)^{5}}{\sqrt{x^{2} + 1}}$.

2. 02

   At $\frac{3}{4}$: $\sqrt{x^{2} + 1} = \frac{5}{4}$ and $x + \sqrt{x^{2} + 1} = 2$, so $f' = \frac{5 \times 32}{5/4} = 128$.