find $\frac{dy}{dx}$ for $y = sqrt{u}$ and $u=x^{2}+4$. state your answer in terms of $x$ only.\n$\frac{dy}{d…

Explanation:

Step1: Find $\frac{dy}{du}$

Since $y = \sqrt{u}=u^{\frac{1}{2}}$, by the power - rule $\frac{d}{du}(u^n)=nu^{n - 1}$, we have $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}$.

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

Given $u=x^{2}+4$, then $\frac{du}{dx} = 2x$.

Step3: Use the chain - rule

The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=\frac{1}{2\sqrt{u}}$ and $\frac{du}{dx}=2x$ into the chain - rule formula. We get $\frac{dy}{dx}=\frac{1}{2\sqrt{u}}\cdot2x=\frac{x}{\sqrt{u}}$.

Step4: Substitute $u = x^{2}+4$ back in

Replace $u$ with $x^{2}+4$ in the expression for $\frac{dy}{dx}$. So $\frac{dy}{dx}=\frac{x}{\sqrt{x^{2}+4}}$.

Answer:

$\frac{x}{\sqrt{x^{2}+4}}$