Derivation of a Derivative

Hi, mathe experts...

I am looking at a calculus intoduction and I am not sure how something was derived, I am pretty good at re-arranging normal formulas but a bit stuck here.

How do we get from:

P = C/(1+r) + C/(1+r)^2 + C/(1+r)^3 + ... + C/(1+r)^9 + C/(1+r)^9 + par/(1+r)^10


dP/dr = -C/(1+r)^2 + -2C/(1+r)^3 + ... -10C/(1+r)^11 + -10 par/(1+r)^11

I know thas the diff of x^2 = 2x
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Since P is a sum, you can differentiate every term separately. They are all of the form :

        C / (1 + r)^n

This is the same as :

        C * (1 + r)^(-n)

The derivative of that is :

        C * (-n) * (1 + r)^(-n - 1) * d(1+r)/dr = -C * n / (1 + r)^(n + 1)

So, for the second term (n = 2) :


we get the derivative :

        -2C / (1 + r)^3

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
sublimationAuthor Commented:
That is plush!
btw :

>> I know thas the diff of x^2 = 2x

This applies the exact same rule :

        d (x^n) / dx = n * x^(n - 1)
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Math / Science

From novice to tech pro — start learning today.