Question

attempt 1: 10 attempts remaining. find $\frac{dy}{dx}$ if $y = (-8x^{3}+3x^{2}-x)^{3}$. $\frac{dy}{dx}=$

Explanation:

Step1: Apply chain - rule

Let $u=-8x^{3}+3x^{2}-x$, then $y = u^{3}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.
First, find $\frac{dy}{du}$. Since $y = u^{3}$, by the power rule $\frac{dy}{du}=3u^{2}$.

Step2: Find $\frac{du}{dx}$

Differentiate $u=-8x^{3}+3x^{2}-x$ with respect to $x$. Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $\frac{du}{dx}=-24x^{2}+6x - 1$.

Step3: Calculate $\frac{dy}{dx}$

Substitute $u=-8x^{3}+3x^{2}-x$ and $\frac{dy}{du}=3u^{2}$, $\frac{du}{dx}=-24x^{2}+6x - 1$ into the chain - rule formula.
$\frac{dy}{dx}=3(-8x^{3}+3x^{2}-x)^{2}(-24x^{2}+6x - 1)$

Answer:

$3(-8x^{3}+3x^{2}-x)^{2}(-24x^{2}+6x - 1)$