Question

1. find $\frac{d^{2}y}{dx^{2}}$ for $y=(x^{9}+e^{x}+1)(x^{5}+x + 2)$.

Explanation:

Step1: Apply product - rule for first - derivative

Let $u = x^{9}+e^{x}+1$ and $v=x^{5}+x + 2$. The product - rule states that $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$.
$\frac{du}{dx}=9x^{8}+e^{x}$ and $\frac{dv}{dx}=5x^{4}+1$.
So, $\frac{dy}{dx}=(x^{9}+e^{x}+1)(5x^{4}+1)+(x^{5}+x + 2)(9x^{8}+e^{x})$.

Step2: Expand the first - derivative

Expand $(x^{9}+e^{x}+1)(5x^{4}+1)=5x^{13}+x^{9}+5x^{4}e^{x}+e^{x}+5x^{4}+1$.
Expand $(x^{5}+x + 2)(9x^{8}+e^{x})=9x^{13}+x^{5}e^{x}+9x^{9}+xe^{x}+18x^{8}+2e^{x}$.
Then $\frac{dy}{dx}=(5x^{13}+x^{9}+5x^{4}e^{x}+e^{x}+5x^{4}+1)+(9x^{13}+x^{5}e^{x}+9x^{9}+xe^{x}+18x^{8}+2e^{x})$.
Combine like - terms: $\frac{dy}{dx}=14x^{13}+10x^{9}+18x^{8}+5x^{4}+5x^{4}e^{x}+x^{5}e^{x}+xe^{x}+3e^{x}+1$.

Step3: Apply sum - rule and product - rule for second - derivative

The sum - rule states that if $y = f_1(x)+f_2(x)+\cdots+f_n(x)$, then $\frac{d^{2}y}{dx^{2}}=\frac{d^{2}f_1}{dx^{2}}+\frac{d^{2}f_2}{dx^{2}}+\cdots+\frac{d^{2}f_n}{dx^{2}}$.
For the product terms, use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ again.
$\frac{d}{dx}(14x^{13}) = 182x^{12}$, $\frac{d}{dx}(10x^{9})=90x^{8}$, $\frac{d}{dx}(18x^{8}) = 144x^{7}$, $\frac{d}{dx}(5x^{4})=20x^{3}$.
For $\frac{d}{dx}(5x^{4}e^{x})$, using the product - rule: $(5x^{4}e^{x})^\prime=20x^{3}e^{x}+5x^{4}e^{x}$.
For $\frac{d}{dx}(x^{5}e^{x})$, using the product - rule: $(x^{5}e^{x})^\prime=5x^{4}e^{x}+x^{5}e^{x}$.
For $\frac{d}{dx}(xe^{x})$, using the product - rule: $(xe^{x})^\prime=e^{x}+xe^{x}$.
For $\frac{d}{dx}(3e^{x}) = 3e^{x}$, $\frac{d}{dx}(1)=0$.
Then $\frac{d^{2}y}{dx^{2}}=182x^{12}+90x^{8}+144x^{7}+20x^{3}+20x^{3}e^{x}+5x^{4}e^{x}+5x^{4}e^{x}+x^{5}e^{x}+e^{x}+xe^{x}+3e^{x}$.
Combine like - terms:
$\frac{d^{2}y}{dx^{2}}=182x^{12}+90x^{8}+144x^{7}+20x^{3}+20x^{3}e^{x}+10x^{4}e^{x}+x^{5}e^{x}+xe^{x}+4e^{x}$.

Answer:

$182x^{12}+90x^{8}+144x^{7}+20x^{3}+20x^{3}e^{x}+10x^{4}e^{x}+x^{5}e^{x}+xe^{x}+4e^{x}$