Question

differentiate the function

$y=(3x - 2)^3(4 - x^4)^3$

$\frac{dy}{dx}=(12x-3x^5 - 8 + 2x^4)^3$

Explanation:

Step1: Apply product - rule

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

Step2: Differentiate $u$ using chain - rule

If $u=(3x - 2)^3$, let $t = 3x-2$, then $u = t^3$. By the chain - rule $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}$. $\frac{du}{dt}=3t^2 = 3(3x - 2)^2$ and $\frac{dt}{dx}=3$, so $\frac{du}{dx}=9(3x - 2)^2$.

Step3: Differentiate $v$ using chain - rule

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

Step4: Calculate $y^\prime$

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

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Answer:

$y^\prime=3(3x - 2)^2(4 - x^4)^2(12 + 8x^3-15x^4)$