Question

find the third derivative of (y = xe^{x}+7x^{2}+9).

Explanation:

Step1: Recall product - rule and power - rule

The product - rule is $(uv)' = u'v+uv'$ where if $u = x$ and $v = e^{x}$, then $u'=1$ and $v' = e^{x}$, so $(xe^{x})'=e^{x}+xe^{x}$. The power - rule is $(x^{n})'=nx^{n - 1}$, so $(7x^{2})' = 14x$ and $(9)' = 0$. Then $y'=(xe^{x})'+(7x^{2})'+(9)'=e^{x}+xe^{x}+14x$.

Step2: Find the second - derivative

$y''=(e^{x}+xe^{x}+14x)'=(e^{x})'+(xe^{x})'+(14x)'$. Since $(e^{x})' = e^{x}$, using the product - rule for $(xe^{x})'$ again gives $(xe^{x})'=e^{x}+xe^{x}$, and $(14x)' = 14$. So $y''=e^{x}+e^{x}+xe^{x}+14=2e^{x}+xe^{x}+14$.

Step3: Find the third - derivative

$y'''=(2e^{x}+xe^{x}+14)'=(2e^{x})'+(xe^{x})'+(14)'$. Since $(2e^{x})' = 2e^{x}$, using the product - rule for $(xe^{x})'$ gives $(xe^{x})'=e^{x}+xe^{x}$, and $(14)' = 0$. Then $y'''=2e^{x}+e^{x}+xe^{x}=3e^{x}+xe^{x}$.

Answer:

$3e^{x}+xe^{x}$