Question

differentiate the function. y=(4x - 4)^4(1 - x^3)^4 dy/dx=□

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u=(4x - 4)^4$ and $v=(1 - x^3)^4$.

Step2: Differentiate $u$ using chain - rule

If $u=(4x - 4)^4$, let $t = 4x-4$, then $u = t^4$. By the chain - rule $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}$. $\frac{du}{dt}=4t^3$ and $\frac{dt}{dx}=4$. So $\frac{du}{dx}=4(4x - 4)^3\times4 = 16(4x - 4)^3$.

Step3: Differentiate $v$ using chain - rule

If $v=(1 - x^3)^4$, let $s = 1 - x^3$, then $v = s^4$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}$. $\frac{dv}{ds}=4s^3$ and $\frac{ds}{dx}=-3x^2$. So $\frac{dv}{dx}=4(1 - x^3)^3\times(-3x^2)=-12x^2(1 - x^3)^3$.

Step4: Apply product - rule result

$y^\prime=\frac{dy}{dx}=u^\prime v+uv^\prime$. Substitute $u$, $v$, $u^\prime$, and $v^\prime$:
\[

$$\begin{align*} \frac{dy}{dx}&=16(4x - 4)^3(1 - x^3)^4+(4x - 4)^4\times(-12x^2(1 - x^3)^3)\\ &=(4x - 4)^3(1 - x^3)^3[16(1 - x^3)-12x^2(4x - 4)]\\ &=(4x - 4)^3(1 - x^3)^3(16-16x^3-48x^3 + 48x^2)\\ &=(4x - 4)^3(1 - x^3)^3(16-64x^3 + 48x^2) \end{align*}$$

\]

Answer:

$(4x - 4)^3(1 - x^3)^3(16-64x^3 + 48x^2)$