differentiate the function.\ny = (5x^4 - x + 1)(-x^5 + 8)\ny =

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = 5x^{4}-x + 1$ and $v=-x^{5}+8$.

Step2: Differentiate $u$

Differentiate $u = 5x^{4}-x + 1$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we get $u'=\frac{d}{dx}(5x^{4})-\frac{d}{dx}(x)+\frac{d}{dx}(1)=20x^{3}-1+0 = 20x^{3}-1$.

Step3: Differentiate $v$

Differentiate $v=-x^{5}+8$ with respect to $x$. Using the power - rule, we get $v'=\frac{d}{dx}(-x^{5})+\frac{d}{dx}(8)=-5x^{4}+0=-5x^{4}$.

Step4: Calculate $y'$

Substitute $u$, $u'$, $v$, and $v'$ into the product - rule formula $y'=u'v + uv'$. [ \begin{align*} y'&=(20x^{3}-1)(-x^{5}+8)+(5x^{4}-x + 1)(-5x^{4})\ &=(20x^{3})(-x^{5})+(20x^{3})\times8+(-1)(-x^{5})+(-1)\times8+(5x^{4})(-5x^{4})+(-x)(-5x^{4})+(1)(-5x^{4})\ &=-20x^{8}+160x^{3}+x^{5}-8 - 25x^{8}+5x^{5}-5x^{4}\ &=(-20x^{8}-25x^{8})+(x^{5}+5x^{5})-5x^{4}+160x^{3}-8\ &=-45x^{8}+6x^{5}-5x^{4}+160x^{3}-8 \end{align*} ]

Answer:

$-45x^{8}+6x^{5}-5x^{4}+160x^{3}-8$