f(x)=e^{8x}(x^{2}+9^{x})\nf(x)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = e^{8x}$ and $v=x^{2}+9^{x}$.

Step2: Differentiate $u = e^{8x}$

Using the chain - rule, if $y = e^{g(x)}$, then $y'=g'(x)e^{g(x)}$. For $u = e^{8x}$, $u'=8e^{8x}$.

Step3: Differentiate $v=x^{2}+9^{x}$

The derivative of $x^{2}$ is $2x$ and the derivative of $9^{x}$ is $9^{x}\ln(9)$ (since if $y = a^{x}$, $y'=a^{x}\ln(a)$). So $v'=2x + 9^{x}\ln(9)$.

Step4: Calculate $f'(x)$

By the product - rule $f'(x)=u'v+uv'$. Substitute $u = e^{8x}$, $u'=8e^{8x}$, $v=x^{2}+9^{x}$, and $v'=2x + 9^{x}\ln(9)$ into the formula: [ \begin{align*} f'(x)&=8e^{8x}(x^{2}+9^{x})+e^{8x}(2x + 9^{x}\ln(9))\ &=e^{8x}(8x^{2}+8\cdot9^{x}+2x + 9^{x}\ln(9))\ &=e^{8x}(8x^{2}+2x+9^{x}(8 + \ln(9))) \end{align*} ]

Answer:

$e^{8x}(8x^{2}+2x + 9^{x}(8+\ln(9)))$