Question

differentiate the function.
y = (8x^4 - x + 2)(-x^5 + 2)
y = - 72x^8+6x^5 - 10x^4+64x^3 - 2

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 $v=-x^{5}+2$. First, find $u'$ and $v'$.
$u'=\frac{d}{dx}(8x^{4}-x + 2)=32x^{3}-1$
$v'=\frac{d}{dx}(-x^{5}+2)=-5x^{4}$

Step2: Calculate $y'$ using the product - rule

$y'=(32x^{3}-1)(-x^{5}+2)+(8x^{4}-x + 2)(-5x^{4})$
Expand the first part:
$(32x^{3}-1)(-x^{5}+2)=32x^{3}\times(-x^{5})+32x^{3}\times2-1\times(-x^{5})-1\times2=-32x^{8}+64x^{3}+x^{5}-2$
Expand the second part:
$(8x^{4}-x + 2)(-5x^{4})=8x^{4}\times(-5x^{4})-x\times(-5x^{4})+2\times(-5x^{4})=-40x^{8}+5x^{5}-10x^{4}$

Step3: Combine like terms

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

Answer:

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