find the derivative of the function. y = (5x - 2)^4(9x^2 - 4)^-3 y =

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=(5x - 2)^{4}$ and $v=(9x^{2}-4)^{-3}$.

Step2: Find $u'$ using chain - rule

The chain - rule: if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. For $u=(5x - 2)^{4}$, let $t = 5x-2$, so $u = t^{4}$. Then $\frac{du}{dt}=4t^{3}$ and $\frac{dt}{dx}=5$. So $u'=\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}=4(5x - 2)^{3}\cdot5 = 20(5x - 2)^{3}$.

Step3: Find $v'$ using chain - rule

For $v=(9x^{2}-4)^{-3}$, let $s = 9x^{2}-4$, so $v = s^{-3}$. Then $\frac{dv}{ds}=-3s^{-4}$ and $\frac{ds}{dx}=18x$. So $v'=\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}=-3(9x^{2}-4)^{-4}\cdot18x=-54x(9x^{2}-4)^{-4}$.

Step4: Calculate $y'$

$y'=u'v+uv'$. Substitute $u$, $u'$, $v$, and $v'$: [ \begin{align*} y'&=20(5x - 2)^{3}(9x^{2}-4)^{-3}+(5x - 2)^{4}\cdot(-54x)(9x^{2}-4)^{-4}\ &=(5x - 2)^{3}(9x^{2}-4)^{-4}[20(9x^{2}-4)-54x(5x - 2)]\ &=(5x - 2)^{3}(9x^{2}-4)^{-4}(180x^{2}-80 - 270x^{2}+108x)\ &=(5x - 2)^{3}(9x^{2}-4)^{-4}(- 90x^{2}+108x - 80)\ &=\frac{(5x - 2)^{3}(-90x^{2}+108x - 80)}{(9x^{2}-4)^{4}} \end{align*} ]

Answer:

$\frac{(5x - 2)^{3}(-90x^{2}+108x - 80)}{(9x^{2}-4)^{4}}$