Question

differentiate the function.
y = (4x - 4)^4(1 - x^3)^4
\frac{dy}{dx}=square

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: Find $u^\prime$ using chain - rule

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 $u^\prime = 4(4x - 4)^3\times4=16(4x - 4)^3$.

Step3: Find $v^\prime$ using chain - rule

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 $v^\prime=4(1 - x^3)^3\times(-3x^2)=-12x^2(1 - x^3)^3$.

Step4: Calculate $y^\prime$

$y^\prime=u^\prime v+uv^\prime$.
$y^\prime = 16(4x - 4)^3(1 - x^3)^4-12x^2(4x - 4)^4(1 - x^3)^3$.
Factor out $4(4x - 4)^3(1 - x^3)^3$:
$y^\prime=4(4x - 4)^3(1 - x^3)^3[4(1 - x^3)-3x^2(4x - 4)]$.
Expand the expression inside the brackets:
$y^\prime=4(4x - 4)^3(1 - x^3)^3(4-4x^3-12x^3 + 12x^2)$.
$y^\prime=4(4x - 4)^3(1 - x^3)^3(4-16x^3 + 12x^2)$.
$y^\prime=16(4x - 4)^3(1 - x^3)^3(1 - 4x^3+3x^2)$.

Answer:

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