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}}

\int_{0}^{π} 3 sin (x) d x

\int_{1}^{6} π (x - 1) d x

\frac{d r}{d t} = 6 (t + 1)^{\frac{1}{2}}, r (0) = 0

(3.6) (5.1)

f (x) = 2 e^{x}