Question

1. (exam 2 review #11b) 5 pts find the differential of the function at the indicated number. $f(x)=sqrt{2x^{2}+4}; x = 4$

Explanation:

Step1: Differentiate the function

Let $u = 2x^{2}+4$, then $y=\sqrt{u}=u^{\frac{1}{2}}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$.
$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=4x$.
By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{2}(2x^{2}+4)^{-\frac{1}{2}}\cdot4x=\frac{2x}{\sqrt{2x^{2}+4}}$.

Step2: Evaluate the derivative at $x = 4$

Substitute $x = 4$ into $\frac{dy}{dx}$.
$\frac{dy}{dx}\big|_{x = 4}=\frac{2\times4}{\sqrt{2\times4^{2}+4}}=\frac{8}{\sqrt{32 + 4}}=\frac{8}{\sqrt{36}}=\frac{8}{6}=\frac{4}{3}$.

Step3: Recall the formula for the differential

The differential $dy$ of a function $y = f(x)$ is given by $dy=f^{\prime}(x)dx$.
When $x = 4$ and $f^{\prime}(4)=\frac{4}{3}$, the differential $dy=\frac{4}{3}dx$.

Answer:

$dy=\frac{4}{3}dx$