How do I take the derivative of (cos(x))^3? Arethese right? --> 3(-sin(x))^2 *cos(x)=-3cos(x)(sin(x))^2
not quite, by the chain rule
y = (cos x)^3
dy/dx = 3(cosx)^2 (-sinx)
Just expanding on Reiny's answer, consider
u = cos^3 x
Then y = u^3
so y' = 3u^2 u' = 3cos^2x (-sinx)
You squared the derivative - tsk tsk
To take the derivative of (cos(x))^3, you can use the chain rule, which states that if you have a composite function f(g(x)), the derivative is given by f'(g(x)) multiplied by g'(x).
In this case, let's define f(u) = u^3 and g(x) = cos(x).
To find the derivative of f(g(x)), we need to find the derivatives of f(u) and g(x) separately:
1. Derivative of f(u):
Using the power rule, the derivative of u^3 is 3u^2.
2. Derivative of g(x):
The derivative of cos(x) is -sin(x).
Now, we can apply the chain rule:
f'(g(x)) * g'(x) = 3(cos(x))^2 * (-sin(x))
So, the correct derivative of (cos(x))^3 is 3(cos(x))^2 * (-sin(x)).
Your calculation: 3(-sin(x))^2 *cos(x) = -3cos(x)(sin(x))^2
It seems there may have been a sign error in your calculation. The correct derivative is -3(cos(x))^2 * sin(x).