# Thread: Derivative of this function

1. ## Derivative of this function

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.

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

now use the power rule

3. 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

4. Its good!

5. 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!

6. I get $\displaystyle y-f(3)= f'(3)(x-3)$

7. Originally Posted by youngb11
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.
Originally Posted by e^(i*pi)
$f(x) = 3x^{3/2}$

now use the power rule
I hope it hasn't escaped notice that $\displaystyle \sqrt{3x^3} \neq 3x^{3/2}$

-Dan

8. ## A slight correction

Originally Posted by youngb11
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.
$\sqrt{3x^3} = 3^{1/2} \cdot x^{3/2}$. Now apply the power rule and check it out.