Originally Posted by

**DarkestEvil** Evaluate the derivative of the function at the indicated points:

$\displaystyle y=37-sec^3(2x)$, point $\displaystyle (0,36)$

Does the first equation become $\displaystyle y=[37-sec(2x)]^3$ and then use chain rule? After I do so, I plug in the point to find a value?

__You have to take the 37 out of the brackets. You can't write it like that. You can just ignore the 37 all together since the derivative of a constant is zero. The derivative of your function is the same as the derivative of:__

$\displaystyle y=-sec^3(2x)$

Find an equation of the tangent line to the graph of f at the indicated point.

$\displaystyle f(x)=sin2x$, point $\displaystyle (\pi,0)$

What do I do here? :[

Thanks for the help in advance.