Question

hw 14 - derivatives of logs section 2.9: problem 8 (1 point)
let $f(x)=4x^{6}ln x$.
$f(x)=square$
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Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, $u = 4x^{6}$ and $v=\ln x$.

Step2: Find $u'$

Using the power - rule $\frac{d}{dx}(ax^{n})=anx^{n - 1}$, for $u = 4x^{6}$, we have $u'=\frac{d}{dx}(4x^{6})=4\times6x^{5}=24x^{5}$.

Step3: Find $v'$

The derivative of $\ln x$ with respect to $x$ is $\frac{1}{x}$, so $v'=\frac{1}{x}$.

Step4: Calculate $f'(x)$

By the product - rule $f'(x)=u'v+uv'$. Substitute $u = 4x^{6}$, $u' = 24x^{5}$, $v=\ln x$, and $v'=\frac{1}{x}$ into the formula:
$f'(x)=24x^{5}\ln x+4x^{6}\times\frac{1}{x}=24x^{5}\ln x + 4x^{5}$.

Answer:

$24x^{5}\ln x+4x^{5}$