find the derivative of the following function. y=x³(5 - lnx³) dy/dx = □ (use parentheses to clearly denote…

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = x^{3}$ and $v=5-\ln(x^{3})$. First, find $u^\prime$ and $v^\prime$. The derivative of $u = x^{3}$ with respect to $x$ is $u^\prime=\frac{d}{dx}(x^{3}) = 3x^{2}$ by the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$.

Step2: Find the derivative of $v$

We have $v = 5-\ln(x^{3})$. Using the constant rule $\frac{d}{dx}(c)=0$ for a constant $c$ and the chain - rule for $\ln(x^{3})$. Let $t=x^{3}$, then $\frac{d}{dx}(\ln(x^{3}))=\frac{d}{dt}(\ln(t))\cdot\frac{dt}{dx}$. Since $\frac{d}{dt}(\ln(t))=\frac{1}{t}$ and $\frac{dt}{dx}=3x^{2}$, $\frac{d}{dx}(\ln(x^{3}))=\frac{3x^{2}}{x^{3}}=\frac{3}{x}$. So $v^\prime=\frac{d}{dx}(5)-\frac{d}{dx}(\ln(x^{3}))=0 - \frac{3}{x}=-\frac{3}{x}$.

Step3: Apply the product - rule formula

$y^\prime=u^\prime v+uv^\prime$. Substitute $u = x^{3}$, $u^\prime = 3x^{2}$, $v = 5-\ln(x^{3})$, and $v^\prime=-\frac{3}{x}$ into the formula: [ \begin{align*} y^\prime&=3x^{2}(5 - \ln(x^{3}))+x^{3}\left(-\frac{3}{x}\right)\ &=15x^{2}-3x^{2}\ln(x^{3})- 3x^{2}\ &=12x^{2}-3x^{2}\ln(x^{3}) \end{align*} ]

Answer:

$12x^{2}-3x^{2}\ln(x^{3})$