Question

differentiate the function.
y = (8x^4 - x + 2)(-x^5 + 2)
y =

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = 8x^{4}-x + 2$ and $v=-x^{5}+2$.

Step2: Differentiate $u$

Differentiate $u = 8x^{4}-x + 2$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we get $u'=\frac{d}{dx}(8x^{4})-\frac{d}{dx}(x)+\frac{d}{dx}(2)=32x^{3}-1+0 = 32x^{3}-1$.

Step3: Differentiate $v$

Differentiate $v=-x^{5}+2$ with respect to $x$. Using the power - rule, we get $v'=\frac{d}{dx}(-x^{5})+\frac{d}{dx}(2)=-5x^{4}+0=-5x^{4}$.

Step4: Substitute $u$, $u'$, $v$, and $v'$ into the product - rule

$y'=(32x^{3}-1)(-x^{5}+2)+(8x^{4}-x + 2)(-5x^{4})$.
Expand the two products:
First product: $(32x^{3}-1)(-x^{5}+2)=32x^{3}\times(-x^{5})+32x^{3}\times2-1\times(-x^{5})-1\times2=-32x^{8}+64x^{3}+x^{5}-2$.
Second product: $(8x^{4}-x + 2)(-5x^{4})=8x^{4}\times(-5x^{4})-x\times(-5x^{4})+2\times(-5x^{4})=-40x^{8}+5x^{5}-10x^{4}$.

Step5: Combine like terms

$y'=(-32x^{8}+64x^{3}+x^{5}-2)+(-40x^{8}+5x^{5}-10x^{4})$.
$y'=(-32x^{8}-40x^{8})+(x^{5}+5x^{5})-10x^{4}+64x^{3}-2$.
$y'=-72x^{8}+6x^{5}-10x^{4}+64x^{3}-2$.

Answer:

$-72x^{8}+6x^{5}-10x^{4}+64x^{3}-2$