find the derivative of the function f(x), below. it may be to your advantage to simplify first. f(x)=(x^9…

Explanation:

Step1: Apply the product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=x^{9}-\sqrt[9]{x}=x^{9}-x^{\frac{1}{9}}$ and $v = 6^{x}$.

Step2: Find the derivative of $u$

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $u'=\frac{d}{dx}(x^{9}-x^{\frac{1}{9}})=9x^{8}-\frac{1}{9}x^{-\frac{8}{9}}$.

Step3: Find the derivative of $v$

The derivative of $a^{x}$ with respect to $x$ is $a^{x}\ln a$. So, $v'=\frac{d}{dx}(6^{x})=6^{x}\ln 6$.

Step4: Apply the product - rule formula

$f'(x)=u'v + uv'=(9x^{8}-\frac{1}{9}x^{-\frac{8}{9}})6^{x}+(x^{9}-x^{\frac{1}{9}})6^{x}\ln 6$. We can factor out $6^{x}$: $f'(x)=6^{x}(9x^{8}-\frac{1}{9x^{\frac{8}{9}}}+(x^{9}-x^{\frac{1}{9}})\ln 6)$.

Answer:

$6^{x}(9x^{8}-\frac{1}{9x^{\frac{8}{9}}}+(x^{9}-x^{\frac{1}{9}})\ln 6)$