# Evaluating a Derivative at a Given point

• Jun 14th 2009, 11:09 AM
VonNemo19
Evaluating a Derivative at a Given point
Find the exact value of $\displaystyle y'(\frac{\pi}{3})$ for $\displaystyle y=3xtanx$.

I got $\displaystyle y'(\frac{\pi}{3})=12+3\sqrt{3}$.

Is that right?

If you would show me how you arrived at your reslut I would appreciate it.
• Jun 14th 2009, 11:42 AM
HallsofIvy
Quote:

Originally Posted by VonNemo19
Find the exact value of $\displaystyle y'(\frac{\pi}{3})$ for $\displaystyle y=3xtanx$.

I got $\displaystyle y'(\frac{\pi}{3})=12+3\sqrt{3}$.

Is that right?

If you would show me how you arrived at your reslut I would appreciate it.

Again, you won't show your work? No, that is not correct.
• Jun 14th 2009, 12:03 PM
VonNemo19
Quote:

Originally Posted by HallsofIvy
Again, you won't show your work? No, that is not correct.

$\displaystyle y'=3x*sec^2x+3*tanx$

$\displaystyle y'(\frac{\pi}{3})=3*\frac{\pi}{3}*\frac{1}{[cos(\frac{\pi}{3})]^2}+3*tan(\frac{\pi}{3})$

$\displaystyle =\pi*4+3*\sqrt{3}$

$\displaystyle =4\pi+3\sqrt{3}$

It looks as if I may have found my problem for this one.

Is this right?
• Jun 14th 2009, 01:27 PM
Jhevon
Quote:

Originally Posted by VonNemo19
$\displaystyle y'=3x*sec^2x+3*tanx$

$\displaystyle y'(\frac{\pi}{3})=3*\frac{\pi}{3}*\frac{1}{[cos(\frac{\pi}{3})]^2}+3*tan(\frac{\pi}{3})$

$\displaystyle =\pi*4+3*\sqrt{3}$

$\displaystyle =4\pi+3\sqrt{3}$

It looks as if I may have found my problem for this one.

Is this right?

yup (Yes)