Find f'(x)
f (x) = q
30 + sin2
(x)
f'(x) = 2cos(x)
f (x) = q^30 + sin^2(x)
I assume a typo, since q is just a constant. So
f'(x) = 2 sinx cosx
I apologize for the typo. Yes, you are correct that the derivative of f(x) = q^30 + sin^2(x) is f'(x) = 2sin(x)cos(x).
To find the derivative of f(x) = 30 + sin^2(x), we need to apply the chain rule.
Step 1: Identify the inner function:
The inner function in this case is g(x) = sin^2(x).
Step 2: Find the derivative of the inner function:
g'(x) = 2sin(x) * cos(x) [using the chain rule]
Step 3: Find the derivative of the outer function:
f'(x) = derivative of 30 (which is 0) + derivative of the inner function
Step 4: Combine the results:
f'(x) = 0 + 2sin(x) * cos(x)
Therefore, f'(x) = 2sin(x) * cos(x).