given the function $y = sqrt{x} cos x$, find $\frac{dy}{dx}$ in any form.\nanswer\n$\frac{dy}{dx}=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, $u = \sqrt{x}=x^{\frac{1}{2}}$ and $v=\cos x$.

Step2: Find $\frac{du}{dx}$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $u = x^{\frac{1}{2}}$, we have $\frac{du}{dx}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$.

Step3: Find $\frac{dv}{dx}$

The derivative of $\cos x$ with respect to $x$ is $-\sin x$, so $\frac{dv}{dx}=-\sin x$.

Step4: Substitute into product - rule

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}=x^{\frac{1}{2}}(-\sin x)+\cos x\cdot\frac{1}{2}x^{-\frac{1}{2}}$. Simplify to get $\frac{dy}{dx}=-\sqrt{x}\sin x+\frac{\cos x}{2\sqrt{x}}$.

Answer:

$-\sqrt{x}\sin x+\frac{\cos x}{2\sqrt{x}}$