How do you find the derivative of #1/(1- x)#?

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Aug 26, 2016

Answer:

#dy/dx = 1/((1-x)^2)#

Explanation:

#y = 1/(1-x) = (1-x)^-1#

If u(x) is a function of x, then by the Power Rule ,

#(du^m)/(dx) = m*u^(m-1) * (du)/(dx)#

#dy/dx = (-1)* (1 - x)^-2* (-1) = 1/((1-x)^2)#

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Rhys Share
Dec 15, 2017

Answer:

#(dy)/(dx)= sum_(r=0)^oo rx^(r-1) , |x|<1 #

Explanation:

I wanted to provide another alternate means of thinking of this:

We must somehow find, #1/(1-x) # in some other way:

We can consider the binomial expansion:

#(1+x)^n = 1 + nx + (n(n-1)x^2)/(2!) + (n(n-1)(n-2)x^3)/(3!) + ...#

for #|x|<1#

#=> #

# (1-x)^n = 1 -nx + (n(n-1)x^2)/(2!) - (n(n-1)(n-2)x^3)/(3!) + ...#

Letting #n=-1# :

#=> (1-x)^(-1) = 1 + x + x^2 + x^3 + ... #

for #|x| < 1 #

So our problem becomes:

#d/(dx) (1 + x + x^2 + x^3 + ...) -= d/(dx) (sum_(r=0) ^oo x ^r )#

#=>#

#1+ 2x + 3x^2 + 4x^3 + ...# = #sum_(r=0)^oo rx^(r-1) #

#(dy)/(dx)= sum_(r=0)^oo rx^(r-1) , |x|<1 #

We can also check this via inputing #1/(1-x)^2 # into the binomial expansion!

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