f(x)=8cos4x
What is the derivative? Im doing online homework and i keep getting an incorrect answer. How do you work out this problem correctly? thanks
you must know that if
y = a cos kx, where k is a constant, then
dy/dx = -a(k)sin kx
so f(x) = 8cos 4x
f'(x) = -32sin 4x
To find the derivative of f(x) = 8cos(4x), we can use the chain rule. The chain rule states that if you have a composite function like f(g(x)), the derivative is found by multiplying the derivative of the outer function (f'(g(x))) by the derivative of the inner function (g'(x)).
In this case, the outer function is f(u) = 8cos(u), where u = 4x, and the inner function is g(x) = 4x.
Let's find the derivatives of f(u) and g(x) separately and then apply the chain rule:
1. Derivative of f(u):
The derivative of cos(u) is -sin(u). So, f'(u) = -8sin(u).
2. Derivative of g(x):
The derivative of 4x with respect to x is simply 4.
Now, we can apply the chain rule by multiplying the derivatives together:
f'(x) = f'(u) * g'(x)
= -8sin(u) * 4
But remember, u = 4x, so substituting the value of u back into the equation:
f'(x) = -8sin(4x) * 4
= -32sin(4x)
Therefore, the derivative of f(x) = 8cos(4x) is f'(x) = -32sin(4x).