Need quick calculus help!?

Use the product rule for this.

Let

A = first factor = 8x^4 - x + 5

dA/dx = 32x^3 - 1

B = second factor = -x^5 +6

dB/dx = -5x^4

Applying the product rule,

dy/dx = A(dB/dx) + B(dA/dx)

and substituting appropriate values,

dy/dx = (8x^4 - x + 5)(-5x^4) + (-x^5 + 6)(32x^3 - 1)

I will stop my actual solution at this point. From here, I trust that you can proceed on your own with the algebraic portion of the solution. I have done the calculus portion so I will let you take over the algebra.

Hope this helps.

What you should do, is use the product rule

f'(x) = uv' + vu'

where:

u = 8x^4-x+5 v= -x^5+6

u' = 4(8)x^(4-1) v'= (5)x^(5-1)

=32x^3 = 5x^4

Therefore, we sub those into our product rule to get this:

f'(x) = (8x^4-x+5)(5x^4) + (-x^5+6)(32x^3)

You can simplify this if you wish. Hope this helped.

y=(8x^4-x+5)(-x^5+6)

dy/dx = (8x^4-x+5)(-5x^4) +((32 x^3 -1)(-x^5 + 6)

= - 72 x^8 + 6 x^5 - 25 x^4 + 192 x^3 - 6

y = - 8x^9 + 48 x^4 + x^6 - 6x - 5x^5 + 30 <=>

y = -8x^9 + x^6 -5x^5 + 48x^4 - 6x + 30