[SOLVED] derivatives

Quote:

Originally Posted by obsmith08

The derivative of (1 + (x^2 + 1)^1/2)^1/2

I'm not really sure how to start this problem out... I see that the chain rule applies.... Would (1 + (x^2 + 1)^1/2) be considered g(x)? Can you show me the steps, please?

You will need to use the chain rule more than once:

$\frac d{dx}\left[1 + (x^2 + 1)^{1/2}\right]^{1/2}$

$=\frac12\left[1 + (x^2 + 1)^{1/2}\right]^{-1/2}\cdot\frac d{dx}\left[1 + (x^2 + 1)^{1/2}\right]$

$=\frac12\left[1 + (x^2 + 1)^{1/2}\right]^{-1/2}\cdot\frac12\left(x^2 + 1\right)^{-1/2}\cdot\frac d{dx}\left[x^2+1\right]$

$=\frac12\left[1 + (x^2 + 1)^{1/2}\right]^{-1/2}\cdot\frac12\left(x^2 + 1\right)^{-1/2}\cdot2x$

Now you can clean it up.