Question

find the derivative of the function. f(x)=(2x - 2)^4(x^2 + x + 1)^5 f(x)= find the derivative of the function. h(t)=(t + 1)^(2/3)(3t^2 - 5)^3 h(t)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=(2x - 2)^{4}$ and $v=(x^{2}+x + 1)^{5}$.

Step2: Find $u'$ using chain - rule

If $u=(2x - 2)^{4}$, let $t = 2x-2$, then $u = t^{4}$. By the chain - rule $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}$. $\frac{du}{dt}=4t^{3}=4(2x - 2)^{3}$ and $\frac{dt}{dx}=2$. So $u'=8(2x - 2)^{3}$.

Step3: Find $v'$ using chain - rule

If $v=(x^{2}+x + 1)^{5}$, let $s=x^{2}+x + 1$, then $v = s^{5}$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}$. $\frac{dv}{ds}=5s^{4}=5(x^{2}+x + 1)^{4}$ and $\frac{ds}{dx}=2x + 1$. So $v'=5(2x + 1)(x^{2}+x + 1)^{4}$.

Step4: Calculate $f'(x)$

$f'(x)=u'v+uv'=8(2x - 2)^{3}(x^{2}+x + 1)^{5}+5(2x + 1)(2x - 2)^{4}(x^{2}+x + 1)^{4}$
$=(2x - 2)^{3}(x^{2}+x + 1)^{4}[8(x^{2}+x + 1)+5(2x + 1)(2x - 2)]$
$=(2x - 2)^{3}(x^{2}+x + 1)^{4}(8x^{2}+8x + 8+5(4x^{2}-4x + 2x - 2))$
$=(2x - 2)^{3}(x^{2}+x + 1)^{4}(8x^{2}+8x + 8 + 20x^{2}-10x - 10)$
$=(2x - 2)^{3}(x^{2}+x + 1)^{4}(28x^{2}-2x - 2)$
$=2(2x - 2)^{3}(x^{2}+x + 1)^{4}(14x^{2}-x - 1)$

For $h(t)=(t + 1)^{\frac{2}{3}}(3t^{2}-5)^{3}$:

Step1: Apply product - rule

Let $u=(t + 1)^{\frac{2}{3}}$ and $v=(3t^{2}-5)^{3}$. Then $h'(t)=u'v+uv'$.

Step2: Find $u'$ using chain - rule

If $u=(t + 1)^{\frac{2}{3}}$, let $m=t + 1$, then $u = m^{\frac{2}{3}}$. $\frac{du}{dm}=\frac{2}{3}m^{-\frac{1}{3}}=\frac{2}{3}(t + 1)^{-\frac{1}{3}}$ and $\frac{dm}{dt}=1$. So $u'=\frac{2}{3}(t + 1)^{-\frac{1}{3}}$.

Step3: Find $v'$ using chain - rule

If $v=(3t^{2}-5)^{3}$, let $n = 3t^{2}-5$, then $v=n^{3}$. $\frac{dv}{dn}=3n^{2}=3(3t^{2}-5)^{2}$ and $\frac{dn}{dt}=6t$. So $v'=18t(3t^{2}-5)^{2}$.

Step4: Calculate $h'(t)$

$h'(t)=\frac{2}{3}(t + 1)^{-\frac{1}{3}}(3t^{2}-5)^{3}+18t(t + 1)^{\frac{2}{3}}(3t^{2}-5)^{2}$
$=(t + 1)^{-\frac{1}{3}}(3t^{2}-5)^{2}[\frac{2}{3}(3t^{2}-5)+18t(t + 1)]$
$=(t + 1)^{-\frac{1}{3}}(3t^{2}-5)^{2}(2t^{2}-\frac{10}{3}+18t^{2}+18t)$
$=(t + 1)^{-\frac{1}{3}}(3t^{2}-5)^{2}(20t^{2}+18t-\frac{10}{3})$
$=\frac{1}{3}(t + 1)^{-\frac{1}{3}}(3t^{2}-5)^{2}(60t^{2}+54t - 10)$

Answer:

$f'(x)=2(2x - 2)^{3}(x^{2}+x + 1)^{4}(14x^{2}-x - 1)$
$h'(t)=\frac{1}{3}(t + 1)^{-\frac{1}{3}}(3t^{2}-5)^{2}(60t^{2}+54t - 10)$