using the chain rule
so make so
now
and
Question # 2 lol. I don't know why I'm having such a hard time getting this....
Determine the derivative of the following fuction:
g(x) = (cosx)^3
I started by using the power rule to come up with:
g '(x) = 3(cosx)^2
= 3(-sinx)^2
According to my book this is wrong though... it has:
g '(x) = (3(cosx)^2)(-sinx)
= -3cos^2xsinx
Which makes no sense to me. Thanks again.
This technique only works if a is a constant.
I find it easiest to remember this:
if f(x) =[g(x)]^a, then
f'(x) = a*[g(x)]^(a-1)*g'(x)
Here's an examples:
f(x) = (3x)^3
f'(x) = 3*(3x)^2 *3 = 81x^2
Another example:
f(x) = (x^2)^3
f'(x) = 3*(x^2)^2*2x = 6x^5
See how it works? So applying it to your problem:
f(x) = [cos(x)]^3
f'(x) = 3* [cos(x)]^2 * (-sin(x))
= -3 cos^2(x) sin(x)