find the derivative of the function. y = (2x - 3)^4(4x^2 - 5)^-3 y = need help? read it watch it submit answer

Explanation:

Step1: Apply product - rule

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

Step2: Find $u'$ using chain - rule

The chain - rule: if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. For $u=(2x - 3)^{4}$, let $t = 2x-3$, so $u = t^{4}$. Then $\frac{du}{dt}=4t^{3}$ and $\frac{dt}{dx}=2$. So $u'=\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}=4(2x - 3)^{3}\cdot2=8(2x - 3)^{3}$.

Step3: Find $v'$ using chain - rule

For $v=(4x^{2}-5)^{-3}$, let $s = 4x^{2}-5$, so $v = s^{-3}$. Then $\frac{dv}{ds}=-3s^{-4}$ and $\frac{ds}{dx}=8x$. So $v'=\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}=-3(4x^{2}-5)^{-4}\cdot8x=-24x(4x^{2}-5)^{-4}$.

Step4: Calculate $y'$

$y'=u'v+uv'$. Substitute $u$, $u'$, $v$, and $v'$: [ \begin{align*} y'&=8(2x - 3)^{3}(4x^{2}-5)^{-3}+(2x - 3)^{4}\cdot(-24x)(4x^{2}-5)^{-4}\ &=\frac{8(2x - 3)^{3}}{(4x^{2}-5)^{3}}-\frac{24x(2x - 3)^{4}}{(4x^{2}-5)^{4}}\ &=\frac{8(2x - 3)^{3}(4x^{2}-5)-24x(2x - 3)^{4}}{(4x^{2}-5)^{4}}\ &=\frac{(2x - 3)^{3}[8(4x^{2}-5)-24x(2x - 3)]}{(4x^{2}-5)^{4}}\ &=\frac{(2x - 3)^{3}(32x^{2}-40-48x^{2}+72x)}{(4x^{2}-5)^{4}}\ &=\frac{(2x - 3)^{3}(-16x^{2}+72x - 40)}{(4x^{2}-5)^{4}}\ &=\frac{-8(2x - 3)^{3}(2x^{2}-9x + 5)}{(4x^{2}-5)^{4}} \end{align*} ]

Answer:

$\frac{-8(2x - 3)^{3}(2x^{2}-9x + 5)}{(4x^{2}-5)^{4}}$