# Derivative of Secant and Sine (w/ Chain Rule)

• Oct 20th 2009, 03:58 PM
DarkestEvil
Derivative of Secant and Sine (w/ Chain Rule)
Evaluate the derivative of the function at the indicated points:
$y=37-sec^3(2x)$, point $(0,36)$

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

Find an equation of the tangent line to the graph of f at the indicated point.
$f(x)=sin2x$, point $(\pi,0)$

What do I do here? :[

Thanks for the help in advance.
• Oct 20th 2009, 04:09 PM
Quote:

Originally Posted by DarkestEvil
Evaluate the derivative of the function at the indicated points:
$y=37-sec^3(2x)$, point $(0,36)$

Does the first equation become $y=[37-sec(2x)]^3$ 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:

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

Find an equation of the tangent line to the graph of f at the indicated point.
$f(x)=sin2x$, point $(\pi,0)$

What do I do here? :[

Thanks for the help in advance.

(Hi)
• Oct 20th 2009, 04:12 PM
Find an equation of the tangent line to the graph of f at the indicated point.
http://www.mathhelpforum.com/math-he...932d15f3-1.gif, point http://www.mathhelpforum.com/math-he...7336d422-1.gif

In general, the equation of a tangent is given by $y-f(a)=f'(a)(x-a)$ where "a" is the x-coordinate of the point through which the tangent passes. In your case this would be $\pi$. It should be easy to apply this formula.
• Oct 20th 2009, 04:38 PM
DarkestEvil
So for the first one, my derivative became: $y'=3[-sec(2x)]^2[2 sec(2x) tan(2x)]$
• Oct 20th 2009, 05:31 PM
So for the first one, my derivative became: $y'=3[-sec(2x)]^2[2 sec(2x) tan(2x)]$
Yes, so now just find the value of the derivative at the point you were given, and that is the slope $f'(a)$.