Question

given $f(x)=(4x^{3}-8x^{2}+3x - 6)^{20}$, find $f(x)$. $f(x)=$

Explanation:

Step1: Identify the outer - function and inner - function

Let $u = 4x^{3}-8x^{2}+3x - 6$, then $y = u^{20}$.

Step2: Differentiate the outer - function with respect to $u$

Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $\frac{dy}{du}=20u^{19}$.

Step3: Differentiate the inner - function with respect to $x$

$\frac{du}{dx}=\frac{d}{dx}(4x^{3}-8x^{2}+3x - 6)=12x^{2}-16x + 3$.

Step4: Apply the chain rule

The chain rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substituting $u = 4x^{3}-8x^{2}+3x - 6$, $\frac{dy}{du}=20u^{19}$, and $\frac{du}{dx}=12x^{2}-16x + 3$ into the chain - rule formula, we get $f^{\prime}(x)=20(4x^{3}-8x^{2}+3x - 6)^{19}(12x^{2}-16x + 3)$.

Answer:

$20(4x^{3}-8x^{2}+3x - 6)^{19}(12x^{2}-16x + 3)$