What is the (-1/((1+x)^2))^' ?

The (-1/((1+x)^2))^' is 2/((1+x)^3)

What is the first (-1/((1+x)^2))^' ?

The first (-1/((1+x)^2))^' is 2/((1+x)^3)

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

\frac{2}{( 1 + x ) ^{3}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{2}{( 1 + x ) ^{3}} (1 + x)^{'}

= - \frac{2}{( 1 + x ) ^{3}} (1 + x)^{'}

(1 + x)^{'} = 1

= - (- \frac{2}{( 1 + x ) ^{3}} \cdot 1)

Simplify

= \frac{2}{( 1 + x ) ^{3}}

\int \frac{a + b x ^{2}}{3 a x + b x ^{3}} d x

\frac{d}{d x} (arccos (3 x))

t \frac{d y}{d t} + 21 y = \frac{1}{t}

\int_{0}^{x} (t^{3} - t)^{5} d t

10 + \frac{50 x}{2 x ^{2} + 9}