Question

find $\frac{dy}{dx}$ if $y = (-8x^{3}+5x^{2}-x)^{5}$. $\frac{dy}{dx}=$

Explanation:

Step1: Apply chain - rule

Let $u=-8x^{3}+5x^{2}-x$, then $y = u^{5}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.

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

Differentiate $y = u^{5}$ with respect to $u$. Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we get $\frac{dy}{du}=5u^{4}$.

Step3: Find $\frac{du}{dx}$

Differentiate $u=-8x^{3}+5x^{2}-x$ with respect to $x$. Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $\frac{du}{dx}=-24x^{2}+10x - 1$.

Step4: Calculate $\frac{dy}{dx}$

Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$ into the chain - rule formula: $\frac{dy}{dx}=5u^{4}\cdot(-24x^{2}+10x - 1)$. Replace $u=-8x^{3}+5x^{2}-x$ back into the equation, so $\frac{dy}{dx}=5(-8x^{3}+5x^{2}-x)^{4}(-24x^{2}+10x - 1)$.

Answer:

$5(-8x^{3}+5x^{2}-x)^{4}(-24x^{2}+10x - 1)$