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}})^{'}

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 x \cdot \frac{e ^{x}}{2 x} d x

\frac{d}{d x} (x e^{3 z (x)})

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

δ (t)

y^{^{'}} - 2 y = 5