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UrbanXrisis
Jan24-05, 09:28 PM
p is a real number
f(x)=1+2p(x-1)+(x-1)^2 for x=<1

My book says that the derivative of this function is:
f'(x)=2p+2(x-1)

Shouldnt it be 2p+2(x-1)+2(x-1)

since the derivative of 2p(x-1) is 2p+2(x-1)
and the derivative of (x-1)^2 is 2(x-1)

Parth Dave
Jan24-05, 09:36 PM
p is a real number. It is constant.

Curious3141
Jan24-05, 10:26 PM
since the derivative of 2p(x-1) is 2p+2(x-1)

You're misapplying the product rule here.

Nx2
Jan24-05, 10:40 PM
this is what i got

f(x)=1+2p(x-1)+(x-1)^2 for x=<1
f'(x)= 0(x-1) + 2p(1) + 2(x-1)(1)
= 2p + 2(x-1)

use product rule and chain rule
hope that helps...

- Tu

Nx2
Jan24-05, 10:44 PM
the derivitive of 2p(x-1) is:

0(x-1) + 2p(1) which gives u 2p using the product rule

note that p is a constant so when u take the derivitive of it its just a plain 0... sorry i suck at explaining...

- Tu