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

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

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(x/(x+1))^'

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

x^{'} = 1

(x + 1)^{'} = 1

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

1 \cdot (x + 1) - 1 \cdot x = 1

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

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