I need to find the square root of:

$\displaystyle

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

$

The answer I got is:

$\displaystyle

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

$

Is this correct? if not please explain.

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- Dec 31st 2009, 11:01 AMbearhugDerivative of Squareroot
I need to find the square root of:

$\displaystyle

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

$

The answer I got is:

$\displaystyle

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

$

Is this correct? if not please explain. - Dec 31st 2009, 11:25 AMvonflex1
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) - Dec 31st 2009, 11:26 AMbearhug
wouldn't the inner derivative be 2x? since it was x^2

- Dec 31st 2009, 11:27 AMvonflex1
my bad... your answer was correct...

- Dec 31st 2009, 11:28 AMbearhug
great...thank you :)