# Derivative of Squareroot

• December 31st 2009, 12:01 PM
bearhug
Derivative of Squareroot
I need to find the square root of:
$
f(x) = \sqrt{x^2+15}
$

$
f'(x) = \frac{x}{\sqrt{x^2+15}}
$

Is this correct? if not please explain.
• December 31st 2009, 12:25 PM
vonflex1
the answer is actually: f'(x) = x / 2*(sqrt(x^2+15)

because: f(x) = (x^2+15)^(1/2)
so: f'(x) = (1/2)*(x^2+15)^(-1/2)*x <------ the inner derivative
and finally: f'(x) = x / 2*(sqrt(x^2+15)
• December 31st 2009, 12:26 PM
bearhug
wouldn't the inner derivative be 2x? since it was x^2
• December 31st 2009, 12:27 PM
vonflex1