Question

question find the derivative of h(x)=(2 - 2x)^2(3x + 1)^2. provide your answer below: h(x)=□

Explanation:

Step1: Apply product - rule

The product - rule states that if $h(x)=u(x)v(x)$, then $h'(x)=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=(2 - 2x)^{2}$ and $v(x)=(3x + 1)^{2}$.

Step2: Find $u'(x)$ using chain - rule

Let $t = 2-2x$, then $u(t)=t^{2}$. By the chain - rule $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}$. $\frac{du}{dt}=2t$ and $\frac{dt}{dx}=-2$. Substituting $t = 2-2x$ back in, we get $u'(x)=2(2 - 2x)\times(-2)=-4(2 - 2x)$.

Step3: Find $v'(x)$ using chain - rule

Let $s=3x + 1$, then $v(s)=s^{2}$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}$. $\frac{dv}{ds}=2s$ and $\frac{ds}{dx}=3$. Substituting $s = 3x + 1$ back in, we get $v'(x)=2(3x + 1)\times3 = 6(3x + 1)$.

Step4: Calculate $h'(x)$

$h'(x)=u'(x)v(x)+u(x)v'(x)$
$h'(x)=-4(2 - 2x)(3x + 1)^{2}+6(3x + 1)(2 - 2x)^{2}$
Factor out $2(2 - 2x)(3x + 1)$:
$h'(x)=2(2 - 2x)(3x + 1)[-2(3x + 1)+3(2 - 2x)]$
$h'(x)=2(2 - 2x)(3x + 1)(-6x-2 + 6 - 6x)$
$h'(x)=2(2 - 2x)(3x + 1)(4 - 12x)$
$h'(x)=4(1 - x)(3x + 1)(4 - 12x)$

Answer:

$4(1 - x)(3x + 1)(4 - 12x)$