let f(x)=(8x² - 2)⁴(-2x² + 6)¹³ f(x)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=(8x^{2}-2)^{4}$ and $v = (-2x^{2}+6)^{13}$.

Step2: Find $u'$ using chain - rule

The chain - rule: if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. For $u=(8x^{2}-2)^{4}$, let $t = 8x^{2}-2$, so $u = t^{4}$. Then $u'=\frac{du}{dt}\cdot\frac{dt}{dx}$. $\frac{du}{dt}=4t^{3}=4(8x^{2}-2)^{3}$ and $\frac{dt}{dx}=16x$. So $u'=4(8x^{2}-2)^{3}\cdot16x = 64x(8x^{2}-2)^{3}$.

Step3: Find $v'$ using chain - rule

For $v = (-2x^{2}+6)^{13}$, let $s=-2x^{2}+6$, so $v = s^{13}$. Then $v'=\frac{dv}{ds}\cdot\frac{ds}{dx}$. $\frac{dv}{ds}=13s^{12}=13(-2x^{2}+6)^{12}$ and $\frac{ds}{dx}=-4x$. So $v'=13(-2x^{2}+6)^{12}\cdot(-4x)=-52x(-2x^{2}+6)^{12}$.

Step4: Calculate $f'(x)$

Using the product - rule $f'(x)=u'v + uv'$. [ \begin{align*} f'(x)&=64x(8x^{2}-2)^{3}(-2x^{2}+6)^{13}+(8x^{2}-2)^{4}\cdot(-52x)(-2x^{2}+6)^{12}\ &=4x(8x^{2}-2)^{3}(-2x^{2}+6)^{12}[16(-2x^{2}+6)-13(8x^{2}-2)]\ &=4x(8x^{2}-2)^{3}(-2x^{2}+6)^{12}(-32x^{2}+96 - 104x^{2}+26)\ &=4x(8x^{2}-2)^{3}(-2x^{2}+6)^{12}(-136x^{2}+122)\ &=8x(8x^{2}-2)^{3}(-2x^{2}+6)^{12}(-68x^{2}+61) \end{align*} ]

Answer:

$8x(8x^{2}-2)^{3}(-2x^{2}+6)^{12}(-68x^{2}+61)$