What is the first (1/(1-x))^' ?

The first (1/(1-x))^' is 1/((1-x)^2)

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(1/(1-x))^'

(\frac{1}{1 - x})^{^{'}}

\frac{1}{( 1 - x ) ^{2}}

Solution steps

(\frac{1}{1 - x})^{'}

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= ((1 - x)^{- 1})^{^{'}}

Apply the chain rule:

- \frac{1}{( 1 - x ) ^{2}} (1 - x)^{'}

= - \frac{1}{( 1 - x ) ^{2}} (1 - x)^{'}

(1 - x)^{'} = - 1

= - \frac{1}{( 1 - x ) ^{2}} (- 1)

Simplify

= \frac{1}{( 1 - x ) ^{2}}

e^{2 t} sin (j^{2} t)

\frac{\partial}{\partial x} ((9 + 20 x) y^{t a n (x)})

f (x) = - x^{2}, (- 3, - 9)

\int \frac{2}{( x - 2 ) ( x ^{4} + 4 )} d x

x \to - \infty lim (- 3 - x)