question\nfind the derivative of $f(x)=8^{x}(-5x^{3}+2x)$.\nprovide your answer below:\n$f(x)=square$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=8^{x}$ and $v(x)=- 5x^{3}+2x$.

Step2: Find the derivative of $u(x)$

The derivative of $a^{x}$ with respect to $x$ is $a^{x}\ln a$. So, if $u(x)=8^{x}$, then $u'(x)=8^{x}\ln 8$.

Step3: Find the derivative of $v(x)$

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $v(x)=-5x^{3}+2x$, we have $v'(x)=-15x^{2}+2$.

Step4: Substitute into product - rule

$f'(x)=u'(x)v(x)+u(x)v'(x)=8^{x}\ln 8(-5x^{3}+2x)+8^{x}(-15x^{2}+2)$. We can factor out $8^{x}$: $f'(x)=8^{x}\left[\ln 8(-5x^{3}+2x)-15x^{2}+2\right]$.

Answer:

$8^{x}\left[\ln 8(-5x^{3}+2x)-15x^{2}+2\right]$