find f'(x) if f(x)= (cosx)^sinx

To find the derivative of f(x) = (cos(x))^sin(x), we will use the chain rule. The chain rule states that if we have a function of the form g(u(x)), the derivative is given by g'(u(x)) * u'(x).

Let's break down the function step by step:

f(x) = (cos(x))^sin(x)

Step 1: Identify the inner function u(x) and the outer function g(u).

In this case, the inner function is cos(x) and the outer function is u^v, where u = cos(x) and v = sin(x).

Step 2: Find the derivative of the outer function g(u) = u^v.

To do this, we use the property of the exponential function: d/dx (u^v) = v * u^(v-1) * du/dx.

For our case, the derivative of (cos(x))^sin(x) with respect to cos(x) is v * u^(v-1), where v = sin(x) and u = cos(x).

So, d/d(cos(x)) [(cos(x))^sin(x)] = sin(x) * (cos(x))^(sin(x) - 1).

Step 3: Find the derivative of the inner function u(x) = cos(x).

The derivative of cos(x) is given by d/dx [cos(x)] = -sin(x).

Step 4: Apply the chain rule.

Using the chain rule, we multiply the derivative of the outer function by the derivative of the inner function.

f'(x) = (sin(x) * (cos(x))^(sin(x) - 1)) * (-sin(x))

Simplifying further would lead to f'(x) = -sin(x)^2 * (cos(x))^(sin(x) - 1).

Therefore, the derivative of f(x) = (cos(x))^sin(x) is f'(x) = -sin(x)^2 * (cos(x))^(sin(x) - 1).