Derivative of this function

**youngb11** · February 10th 2011, 01:10 PM

How would I find the derivative of this function: $f(x)=\sqrt{3x^3}$ ?

I'm only supposed to use the power rule and I'm having a little trouble.

**e^(i*pi)** · February 10th 2011, 01:12 PM

$f(x) = 3x^{3/2}$

now use the power rule

**youngb11** · February 10th 2011, 01:16 PM

Originally Posted by e^(i*pi)

$f(x) = 3x^{3/2}$

now use the power rule

Thanks! Doing that I got $(9\sqrt{x})/2$ Is that correct?

I entered the original function on wolfram alpha and got a different answer for the derivative:S

**pickslides** · February 10th 2011, 01:22 PM

Its good!

**youngb11** · February 10th 2011, 01:31 PM

Originally Posted by pickslides

Its good!

Thanks! Back to the actual question, however, I had to write an equation for the tangent of $f(x)$ at (3,9)

What I'd normally do is set $f\prime(3)$ to find the slope of the tangent, which would equal to $(9\sqrt{3})/2$ and set that equal to $(y-9)/(x-3)$ to find the equation in standard form.

The back has the final answer as $9x-2y-9=0$ , can anyone tell me if I'm doing anything wrong or if its a mistake in the book? Thanks!