Final answer:
The derivative of the function f(x)=ln(xcosx) is found by applying the product and chain rules of differentiation, resulting in f'(x)= (cosx - xsinx) / (x cosx).
Step-by-step explanation:
To solve the derivative of the function f(x)=ln(xcosx), we apply the product rule and the chain rule of differentiation. First, we take the derivative of the natural logarithm of a product, ln(uv), which is given by:
(ln(uv))' = (1/uv)(u'v + uv')
In this case, u = x and v = cosx. Thus, we differentiate both of them separately:
- u' = 1
- v' = -sinx (derivative of cosx)
Now, we can find the derivative of f(x):
f'(x) = (1/(x cosx)) (1 · cosx + x · (-sinx))
Fully simplify to get the derivative of f(x):
f'(x) = (cosx - xsinx) / (x cosx)
Finally, you may further simplify the expression if required.