elementary derivative problem

[sambellamy]

Hello, I am in third-term calculus and cannot remember a simple derivative problem. I am asked to find what differentiates to:

1/(1+x)²

I rewrote this as (1+x)^-2 , is this an issue? Because when I integrate this, I get

-2(1+x)^-1 , or -2/(1+x). My calculator gives the same answer the book does - that d/dx -1/(1+x) = 1/(1+x)². Can anybody help me out by explaining where the error in my thinking is?

 

[Ishuda]

Hello, I am in third-term calculus and cannot remember a simple derivative problem. I am asked to find what differentiates to:

1/(1+x)²

I rewrote this as (1+x)^-2 , is this an issue? Because when I integrate this, I get

-2(1+x)^-1 , or -2/(1+x). My calculator gives the same answer the book does - that d/dx -1/(1+x) = 1/(1+x)². Can anybody help me out by explaining where the error in my thinking is?

I think you are getting confused between taking derivatives
(xⁿ)' = n x^n-1
and integrating
\(\displaystyle \int x^n dx = \frac{1}{n+1} x^{n+1}, n \ne -1\)

In your case, you have the integral and n=-2 or
\(\displaystyle \int (1+x)^{-2} dx = \frac{1}{-2+1} (1+x)^{-2+1} = - (1 + x)^{-1}\)

Edit to add: Plus, of course, that constant that gets left out every so often.

 

[sambellamy]

Ha, thank you! I was indeed combining parts of the two.