find y and y. y = x^2 ln(7x) y = y =

Explanation:

Step1: Apply product - rule for $y'$

The product - rule states that if $y = uv$, where $u = x^{2}$ and $v=\ln(7x)$, then $y'=u'v + uv'$. First, find $u'$ and $v'$. $u'=\frac{d}{dx}(x^{2}) = 2x$, and $v'=\frac{d}{dx}(\ln(7x))=\frac{1}{x}$. Then $y'=2x\ln(7x)+x^{2}\cdot\frac{1}{x}=2x\ln(7x)+x$.

Step2: Apply product - rule for $y''$

For $y' = 2x\ln(7x)+x$, find the derivative of each term. For the first term $2x\ln(7x)$ (using product - rule with $u = 2x$ and $v=\ln(7x)$), $u' = 2$ and $v'=\frac{1}{x}$, so the derivative of $2x\ln(7x)$ is $2\ln(7x)+2x\cdot\frac{1}{x}=2\ln(7x) + 2$. The derivative of $x$ is $1$. So $y''=2\ln(7x)+2 + 1=2\ln(7x)+3$.

Answer:

$y'=2x\ln(7x)+x$ $y''=2\ln(7x)+3$