Question

differentiate the function.
y=(3x - 2)^3(4 - x^4)^3
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: Find $u^\prime$ using the chain - rule

If $u=(3x - 2)^3$, let $t = 3x-2$, then $u = t^3$. By the chain - rule $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}$. We know that $\frac{du}{dt}=3t^2$ and $\frac{dt}{dx}=3$. Substituting $t = 3x - 2$ back in, we get $u^\prime=3(3x - 2)^2\cdot3 = 9(3x - 2)^2$.

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

If $v=(4 - x^4)^3$, let $s=4 - x^4$, then $v = s^3$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}$. We know that $\frac{dv}{ds}=3s^2$ and $\frac{ds}{dx}=-4x^3$. Substituting $s = 4 - x^4$ back in, we get $v^\prime=3(4 - x^4)^2\cdot(-4x^3)=-12x^3(4 - x^4)^2$.

Step4: Apply the product - rule to find $y^\prime$

$y^\prime=u^\prime v+uv^\prime$. Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the product - rule formula:
\[

$$\begin{align*} y^\prime&=9(3x - 2)^2(4 - x^4)^3+(3x - 2)^3\cdot(-12x^3(4 - x^4)^2)\\ &=(3x - 2)^2(4 - x^4)^2[9(4 - x^4)-12x^3(3x - 2)]\\ &=(3x - 2)^2(4 - x^4)^2(36-9x^4-36x^4 + 24x^3)\\ &=(3x - 2)^2(4 - x^4)^2(36 + 24x^3-45x^4) \end{align*}$$

\]

Answer:

$(3x - 2)^2(4 - x^4)^2(36 + 24x^3-45x^4)$