Question

differentiate the function.
y=(3x - 2)^3(4 - x^4)^3
\frac{dy}{dx}=

Explanation:

Step1: Apply the product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u=(3x - 2)^3$ and $v=(4 - x^4)^3$.

Step2: Differentiate $u$ using the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. For $u=(3x - 2)^3$, let $t = 3x-2$, so $u=t^3$. Then $\frac{du}{dt}=3t^2$ and $\frac{dt}{dx}=3$. So, $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}=3(3x - 2)^2\cdot3 = 9(3x - 2)^2$.

Step3: Differentiate $v$ using the chain - rule

For $v=(4 - x^4)^3$, let $s = 4 - x^4$, so $v=s^3$. Then $\frac{dv}{ds}=3s^2$ and $\frac{ds}{dx}=-4x^3$. So, $\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}=3(4 - x^4)^2\cdot(-4x^3)=-12x^3(4 - x^4)^2$.

Step4: Apply the product - rule result

$y^\prime=\frac{dy}{dx}=\frac{du}{dx}v+u\frac{dv}{dx}=9(3x - 2)^2(4 - x^4)^3+(3x - 2)^3(-12x^3)(4 - x^4)^2$.
Factor out $3(3x - 2)^2(4 - x^4)^2$:
\[

$$\begin{align*} \frac{dy}{dx}&=3(3x - 2)^2(4 - x^4)^2[3(4 - x^4)-4x^3(3x - 2)]\\ &=3(3x - 2)^2(4 - x^4)^2(12-3x^4-12x^4 + 8x^3)\\ &=3(3x - 2)^2(4 - x^4)^2(12 + 8x^3-15x^4) \end{align*}$$

\]

Answer:

$3(3x - 2)^2(4 - x^4)^2(12 + 8x^3-15x^4)$