evaluate the derivative of the following function.\n\nf(x)=\\cos ^{-1}\\left(e^{5 \\cos x}\\right)\n\nf^{prim…

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Let (u = e^{5\cos x}), so (f(x)=\cos^{- 1}(u)). The derivative of (\cos^{-1}(u)) with respect to (u) is (-\frac{1}{\sqrt{1 - u^{2}}}).

Step2: Differentiate the inner function

Differentiate (u = e^{5\cos x}) with respect to (x). Using the chain rule again (if (y = e^{v}) and (v = 5\cos x), then (\frac{dy}{dx}=\frac{dy}{dv}\cdot\frac{dv}{dx})). The derivative of (e^{v}) with respect to (v) is (e^{v}), and the derivative of (v = 5\cos x) with respect to (x) is (- 5\sin x). So (\frac{du}{dx}=e^{5\cos x}\cdot(-5\sin x))

Step3: Combine the results

By the chain rule (f^\prime(x)=-\frac{1}{\sqrt{1-(e^{5\cos x})^{2}}}\cdot(e^{5\cos x}\cdot(- 5\sin x))) Simplify the expression: (f^\prime(x)=\frac{5e^{5\cos x}\sin x}{\sqrt{1 - e^{10\cos x}}})

Answer:

(\frac{5e^{5\cos x}\sin x}{\sqrt{1 - e^{10\cos x}}})