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

Then teach the underlying concepts
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#### Explanation

Explain in detail...

#### Explanation:

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23
Aug 26, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\left(1 - x\right)}^{2}}$

#### Explanation:

$y = \frac{1}{1 - x} = {\left(1 - x\right)}^{-} 1$

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

$\frac{{\mathrm{du}}^{m}}{\mathrm{dx}} = m \cdot {u}^{m - 1} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- 1\right) \cdot {\left(1 - x\right)}^{-} 2 \cdot \left(- 1\right) = \frac{1}{{\left(1 - x\right)}^{2}}$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Rhys Share
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!

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