given the function $y = sqrt4{x}cos x$, find $\frac{dy}{dx}$ in any form.

Explanation:

Step1: Identify the product - rule

The function $y = \sqrt[4]{x}\cos x$ is a product of two functions $u(x)=\sqrt[4]{x}=x^{\frac{1}{4}}$ and $v(x)=\cos x$. The product - rule states that if $y = u(x)v(x)$, then $\frac{dy}{dx}=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$.

Step2: Find the derivative of $u(x)$

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $u(x)=x^{\frac{1}{4}}$, we have $u^{\prime}(x)=\frac{1}{4}x^{\frac{1}{4}-1}=\frac{1}{4}x^{-\frac{3}{4}}$.

Step3: Find the derivative of $v(x)$

The derivative of $v(x)=\cos x$ is $v^{\prime}(x)=-\sin x$.

Step4: Apply the product - rule

Substitute $u(x),u^{\prime}(x),v(x),v^{\prime}(x)$ into the product - rule formula: $\frac{dy}{dx}=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)=\frac{1}{4}x^{-\frac{3}{4}}\cos x+x^{\frac{1}{4}}(-\sin x)$. Simplify to get $\frac{dy}{dx}=\frac{\cos x}{4\sqrt[4]{x^{3}}}-\sqrt[4]{x}\sin x$.

Answer:

$\frac{\cos x}{4\sqrt[4]{x^{3}}}-\sqrt[4]{x}\sin x$