Thread: Derivative of Secant and Sine (w/ Chain Rule)

1. Derivative of Secant and Sine (w/ Chain Rule)

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?

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.

2. 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.

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

In general, the equation of a tangent is given by $\displaystyle y-f(a)=f'(a)(x-a)$ where "a" is the x-coordinate of the point through which the tangent passes. In your case this would be $\displaystyle \pi$. It should be easy to apply this formula.

4. So for the first one, my derivative became: $\displaystyle y'=3[-sec(2x)]^2[2 sec(2x) tan(2x)]$

5. Originally Posted by DarkestEvil
So for the first one, my derivative became: $\displaystyle y'=3[-sec(2x)]^2[2 sec(2x) tan(2x)]$
Yes, so now just find the value of the derivative at the point you were given, and that is the slope $\displaystyle f'(a)$.