Derivative of f(x)=1+2p(x-1)+(x-1)^2

[UrbanXrisis]

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]

p is a real number. It is constant.

 

[Curious3141]

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

You're misapplying the product rule here.

 

[Nx2]

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]

the derivative 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 derivative of it its just a plain 0... sorry i suck at explaining...

- Tu