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

Answer 1

Evelyn Agard

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

I wanted to offer an additional perspective on this:

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

We can think about the expansion of binomials:

#(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 #

Thus, the issue we have is:

#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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Answer 2

Christopher Agresta

To find the derivative of ( \frac{1}{1 - x} ), you can use the chain rule. The derivative is ( \frac{d}{dx} \left( \frac{1}{1 - x} \right) = \frac{d}{dx} \left( (1 - x)^{-1} \right) ). Applying the chain rule, this becomes ( (-1)(1 - x)^{-2}(-1) = \frac{1}{(1 - x)^{2}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.