Quick derivative question.? .R(p)= 1000pe^(-0.02p) Find R'(p) I know how to start it: (1000)(e^(-0.02p)) + (1000p)(__?__) I get stuck there..

By: Guest
Date: Unknown--
Response
0

Answer 1

(-0.02)e^(-0.02p)

Answer 2

If I remember right:

de^x/dx = e^x

function is is own derivative (maybe, check it)

Answer 3

ide like to help ya but i got stuck about here (1000p................)...........?

Answer 4

you need the product rule:

R(p) = 1000p * e^(-0.02p)

1000 p is the first part, and e^(-0.02p) is the second.

derivatives for them are:

1000 and -0.02 e^*(-0.02p) respectively

now, plug them in:

R ' = 1000p * (-0.02 e^(-0.02p)) + e^(-0.02p) *(1000)

= -20p e^(-0.02p)+1000e^(-0.02p)

good luck

[d] By: Guest
Date: Unknown---
Response
What is 1 + 100



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