Derivatives......yuck!?

Question

If y = 1/1-x, find y', y", y''' and guess a formula for d"y/dx"

Answer 1

y = 1/(1-x)
dy/dx = 1/(1-x)²
d²y/dx² = 2/(1-x)³
d³y/dx³ = 6/(1-x)⁴

In general, d^n y/dx^n = n!/(1-x)^(n+1). And I don't have to guess, this is easy to prove by mathematical induction.

Edit: The first poster apparently forgot to apply the chain rule when computing the derivatives. d(1/(1-x))/dx = -1*(1-x)^(-2) * d(1-x)/dx = 1/(1-x)², not -1/(1-x)².

Answer 2

y'= (-1)*(-1)*(1-x)^-2
y''=(-2)*(-1)*(1-x)^-3
y'''=(2)(-3)(-1)(1-x)^-4
You can use either quotient rule or chain rule to get the same answer. It's best that you use chain rule because quotient rule takes more time.

Answer 3

if you like chain rule, rewite as (1-x)^-1 = y
sooooooo ... i won't simplify so you can see:
y'= (-1)*(-1)*(1-x)^-2
y''=(-2)*(-1)*(1-x)^-3
y'''=(2)(-3)(-1)(1-x)^-4

cheers :-Þ

Answer 4

y' = 1/(1-x)^2
y'' = -2/(1-x)^3
y"' = 6/(1-x)^4

d^ny/dx^2 = ((-1)^(n+1) * n!)/((1-x)^(n+1)) where n is the nth derivative of the function.

Had a typo, fixed it. (Had * instead of /)
The (-1)^(n+1) makes the sign of the derivative alternate.
n! is the coefficient.
(1-x)^(n+1) is sort of self-explanatory