Question

find the first and second derivatives. y = √(x + 5) y = 1 / (2√(x + 5)) y =

Explanation:

Step1: Recall quotient - rule for differentiation

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $y'=\frac{1}{2\sqrt{x + 5}}=(x + 5)^{-\frac{1}{2}}$, where $u = 1$, $u'=0$, $v = 2\sqrt{x+5}=2(x + 5)^{\frac{1}{2}}$, and $v'=2\times\frac{1}{2}(x + 5)^{-\frac{1}{2}}=(x + 5)^{-\frac{1}{2}}$.

Step2: Apply the quotient - rule

\[

$$\begin{align*} y''&=\frac{0\times2\sqrt{x + 5}-1\times(x + 5)^{-\frac{1}{2}}}{(2\sqrt{x + 5})^{2}}\\ &=\frac{- (x + 5)^{-\frac{1}{2}}}{4(x + 5)}\\ &=-\frac{1}{4(x + 5)^{\frac{3}{2}}} \end{align*}$$

\]

Answer:

$y''=-\frac{1}{4(x + 5)^{\frac{3}{2}}}$