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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

(x + 1)^{'} = 1

(x - 1)^{'} = 1

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

Simplify

\frac{1 \cdot ( x - 1 ) - 1 \cdot ( x + 1 )}{( x - 1 ) ^{2}} : - \frac{2}{( x - 1 ) ^{2}}

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

\frac{d y}{d t} = 5 y

\int x + x d x

y = \frac{2 x ^{5} + 8}{x ^{3}}

\int \frac{1}{( 2 x ^{2} - x - 3 )} d x

x \to \infty lim (\frac{e ^{2 x}}{e ^{2 x} + 1})