Question

find the derivative of the function.
f(x)=\sqrt{5 + x\sin x}
\frac{d}{dx}\sqrt{5 + x\sin x}=\square

Explanation:

Step1: Rewrite the function

Rewrite \( f(x)=\sqrt{5 + x\sin x}=(5 + x\sin x)^{\frac{1}{2}} \).

Step2: Apply the chain rule

The chain rule states that if \( y = u^{\frac{1}{2}}\) and \(u = 5+x\sin x\), then \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
First, find \(\frac{dy}{du}\): \(\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{5 + x\sin x}}\).
Then, find \(\frac{du}{dx}\) using the sum rule (\((a + b)^\prime=a^\prime + b^\prime\)) and the product rule (\((uv)^\prime = u^\prime v+uv^\prime\) where \(u = x\) and \(v=\sin x\)).
\(\frac{du}{dx}=\frac{d(5)}{dx}+\frac{d(x\sin x)}{dx}\). Since \(\frac{d(5)}{dx} = 0\), and \(\frac{d(x\sin x)}{dx}=\sin x+x\cos x\) (by product rule: \((x)^\prime\sin x+x(\sin x)^\prime=\sin x + x\cos x\)).

Step3: Multiply the two derivatives

\(\frac{dy}{dx}=\frac{1}{2\sqrt{5 + x\sin x}}\cdot(\sin x + x\cos x)\).

Answer:

\(\frac{\sin x + x\cos x}{2\sqrt{5 + x\sin x}}\)