17. 0 / 2 points
find ( y ) and ( y ).
( y=sqrt{x} ln (x) )
( y= )
( y= )

Question

Accepted Answer

# Explanation:
## Step1: Find the first derivative using the product rule
The product rule is $(uv)^\prime = u^\prime v+uv^\prime$. Let $u = \sqrt{x}=x^{\frac{1}{2}}$, then $u^\prime=\frac{1}{2}x^{-\frac{1}{2}}$, and $v = \ln(x)$, $v^\prime=\frac{1}{x}$.
$$
\begin{align*}
y^\prime&=(x^{\frac{1}{2}}\ln(x))^\prime\
&=\frac{1}{2}x^{-\frac{1}{2}}\ln(x)+x^{\frac{1}{2}}\cdot\frac{1}{x}\
&=\frac{\ln(x)}{2\sqrt{x}}+\frac{1}{\sqrt{x}}\
&=\frac{\ln(x) + 2}{2\sqrt{x}}
\end{align*}
$$
## Step2: Find the second derivative using the quotient rule
The quotient rule is $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Let $u=\ln(x)+2$, $u^\prime=\frac{1}{x}$, and $v = 2\sqrt{x}=2x^{\frac{1}{2}}$, $v^\prime=x^{-\frac{1}{2}}$.
$$
\begin{align*}
y^{\prime\prime}&=\frac{\frac{1}{x}\cdot2\sqrt{x}-(\ln(x)+2)\cdot x^{-\frac{1}{2}}}{(2\sqrt{x})^{2}}\
&=\frac{\frac{2}{\sqrt{x}}-\frac{\ln(x)+2}{\sqrt{x}}}{4x}\
&=\frac{\frac{2-(\ln(x)+2)}{\sqrt{x}}}{4x}\
&=\frac{-\ln(x)}{4x\sqrt{x}}
\end{align*}
$$

# Answer:
$y^\prime=\frac{\ln(x)+2}{2\sqrt{x}}$, $y^{\prime\prime}=\frac{-\ln(x)}{4x\sqrt{x}}$

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QUESTION IMAGE

Question

Explanation:

Step1: Find the first derivative using the product rule

Step2: Find the second derivative using the quotient rule

Answer: