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

Dec 15, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sum}_{r = 0}^{\infty} r {x}^{r - 1} , | x | < 1$

#### Explanation:

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

We must somehow find, $\frac{1}{1 - x}$ 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 $| x | < 1$

$\implies$

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

Letting $n = - 1$ :

$\implies {\left(1 - x\right)}^{- 1} = 1 + x + {x}^{2} + {x}^{3} + \ldots$

for $| x | < 1$

So our problem becomes:

$\frac{d}{\mathrm{dx}} \left(1 + x + {x}^{2} + {x}^{3} + \ldots\right) \equiv \frac{d}{\mathrm{dx}} \left({\sum}_{r = 0}^{\infty} {x}^{r}\right)$

$\implies$

$1 + 2 x + 3 {x}^{2} + 4 {x}^{3} + \ldots$ = ${\sum}_{r = 0}^{\infty} r {x}^{r - 1}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sum}_{r = 0}^{\infty} r {x}^{r - 1} , | x | < 1$

We can also check this via inputing $\frac{1}{1 - x} ^ 2$ into the binomial expansion!