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attempt 1: 10 attempts remaining. let $f(x)=5x^{2}ln x$. $f(x)=$

Question

attempt 1: 10 attempts remaining. let $f(x)=5x^{2}ln x$. $f(x)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, $u = 5x^{2}$ and $v=\ln x$.

Step2: Differentiate $u$

Differentiate $u = 5x^{2}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $u'=\frac{d}{dx}(5x^{2})=10x$.

Step3: Differentiate $v$

Differentiate $v=\ln x$ with respect to $x$. We know that $\frac{d}{dx}(\ln x)=\frac{1}{x}$.

Step4: Calculate $f'(x)$

By the product - rule $f'(x)=u'v + uv'$. Substitute $u = 5x^{2}$, $u' = 10x$, $v=\ln x$, and $v'=\frac{1}{x}$ into the formula:
\[

$$\begin{align*} f'(x)&=(10x)\ln x+5x^{2}\cdot\frac{1}{x}\\ &=10x\ln x + 5x \end{align*}$$

\]

Answer:

$10x\ln x + 5x$