Question

find y and y. y = √x ln(x) y = y = resources read it submit answer 12. - / 2 points find y and y. y = ln(7 + ln(x)) y = y = resources

Explanation:

Step1: Find $y'$ for $y = \sqrt{x}\ln(x)$

Use product - rule $(uv)'=u'v + uv'$, where $u = \sqrt{x}=x^{\frac{1}{2}}$ and $v=\ln(x)$.
$u'=\frac{1}{2}x^{-\frac{1}{2}}$ and $v'=\frac{1}{x}$.
$y'=\frac{1}{2}x^{-\frac{1}{2}}\ln(x)+x^{\frac{1}{2}}\cdot\frac{1}{x}=\frac{\ln(x)}{2\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{\ln(x) + 2}{2\sqrt{x}}$

Step2: Find $y''$ for $y = \sqrt{x}\ln(x)$

Use quotient - rule $(\frac{u}{v})'=\frac{u'v - uv'}{v^{2}}$, where $u=\ln(x)+2$ and $v = 2\sqrt{x}=2x^{\frac{1}{2}}$.
$u'=\frac{1}{x}$ and $v'=x^{-\frac{1}{2}}$.
$y''=\frac{\frac{1}{x}\cdot2x^{\frac{1}{2}}-(\ln(x) + 2)\cdot x^{-\frac{1}{2}}}{4x}=\frac{2x^{-\frac{1}{2}}-x^{-\frac{1}{2}}\ln(x)-2x^{-\frac{1}{2}}}{4x}=-\frac{\ln(x)}{4x^{\frac{3}{2}}}$

Step3: Find $y'$ for $y=\ln(7+\ln(x))$

Use chain - rule. Let $u = 7+\ln(x)$, then $y=\ln(u)$.
$\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dx}=\frac{1}{x}$.
$y'=\frac{1}{x(7 + \ln(x))}$

Step4: Find $y''$ for $y=\ln(7+\ln(x))$

Use quotient - rule $(\frac{u}{v})'=\frac{u'v - uv'}{v^{2}}$, where $u = 1$ and $v=x(7+\ln(x))$.
$u' = 0$ and $v'=(7+\ln(x))+x\cdot\frac{1}{x}=8+\ln(x)$.
$y''=\frac{0 - 1\cdot(8+\ln(x))}{x^{2}(7+\ln(x))^{2}}=-\frac{8+\ln(x)}{x^{2}(7+\ln(x))^{2}}$

Answer:

For $y = \sqrt{x}\ln(x)$:
$y'=\frac{\ln(x)+2}{2\sqrt{x}}$
$y''=-\frac{\ln(x)}{4x^{\frac{3}{2}}}$
For $y=\ln(7+\ln(x))$:
$y'=\frac{1}{x(7+\ln(x))}$
$y''=-\frac{8+\ln(x)}{x^{2}(7+\ln(x))^{2}}$