Differentiate $y=\arccos \left(e^{5 x}\right)$.
Answer: $\frac{d y}{d x}=\frac{1}{5} \ln (\cos (x))$

Step 1 ：\(y = \arccos(e^{5x})\)

Step 2 ：Differentiate both sides with respect to \(x\):

Step 3 ：\(\frac{dy}{dx} = \frac{d}{dx} \arccos(e^{5x})\)

Step 4 ：Use the chain rule: \(\frac{dy}{dx} = \frac{d}{de^{5x}} \arccos(e^{5x}) \cdot \frac{d}{dx} e^{5x}\)

Step 5 ：Differentiate \(\arccos(e^{5x})\) with respect to \(e^{5x}\): \(\frac{d}{de^{5x}} \arccos(e^{5x}) = -\frac{1}{\sqrt{1 - (e^{5x})^2}}\)

Step 6 ：Differentiate \(e^{5x}\) with respect to \(x\): \(\frac{d}{dx} e^{5x} = 5e^{5x}\)

Step 7 ：Substitute the derivatives back into the chain rule: \(\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (e^{5x})^2}} \cdot 5e^{5x}\)

Step 8 ：Simplify the expression: \(\frac{dy}{dx} = -\frac{5e^{5x}}{\sqrt{1 - e^{10x}}}\)

Step 9 ：\(\boxed{\frac{dy}{dx} = -\frac{5e^{5x}}{\sqrt{1 - e^{10x}}}}\)