Question

evaluate the derivative of the following function.

f(x)=\cos ^{-1}\left(e^{5 \cos x}\
ight)

f^{prime}(x)=\square

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:

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