Thread: Need help with 2nd derivative

y = t/(t+1)^2

Finding the first derivative:
y' = (t+1)^2 (1) - [t(t+1)(2)]/(t+1)^4
y' = (t+1)[(t+1) - 2t]/(t+1)^4
y' = (1-t)/(t+1)^3

Assuming that is correct, then I'll move on to the second derivative:
y" = (t+1)^3(-1) - [(1-t)(3)(t+1)^2]/(t+1)^6
y" = -(t+1)^2[3(1-t)]/(t+1)^6 I really start having problems here:
y" = (1-t)^2[(3-3t)/(t+1)^6]

Hmmm....could you give me a boost?

Originally Posted by becky

y = t/(t+1)^2

Finding the first derivative:
y' = (t+1)^2 (1) - [t(t+1)(2)]/(t+1)^4
y' = (t+1)[(t+1) - 2t]/(t+1)^4
y' = (1-t)/(t+1)^3

Assuming that is correct, then I'll move on to the second derivative:
y" = (t+1)^3(-1) - [(1-t)(3)(t+1)^2]/(t+1)^6
y" = -(t+1)^2[3(1-t)]/(t+1)^6 I really start having problems here:
y" = (1-t)^2[(3-3t)/(t+1)^6]

Hmmm....could you give me a boost?

First, you really need to put () around the numerator in the first line!

You factored the second line wrong.

y" = {(t+1)^3(-1) - [(1-t)(3)(t+1)^2]}/(t+1)^6

Taking a common (t+1)^2 from the first and second terms gives:
y'' = (t+1)^2{-(t+1) - 3(1-t)}/(t+1)^6

y'' = (t+1)^2{2t-4}/(t+1)^6

y'' = 2(t-2)/(t+1)^4

-Dan