What is the (2/((1+x)^3))^' ?

The (2/((1+x)^3))^' is -6/((1+x)^4)

What is the first (2/((1+x)^3))^' ?

The first (2/((1+x)^3))^' is -6/((1+x)^4)

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

- \frac{6}{( 1 + x ) ^{4}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{3}{( 1 + x ) ^{4}} (1 + x)^{'}

= - \frac{3}{( 1 + x ) ^{4}} (1 + x)^{'}

(1 + x)^{'} = 1

= 2 (- \frac{3}{( 1 + x ) ^{4}} \cdot 1)

Simplify

2 (- \frac{3}{( 1 + x ) ^{4}} \cdot 1) : - \frac{6}{( 1 + x ) ^{4}}

= - \frac{6}{( 1 + x ) ^{4}}

1 - 3 x^{2} \frac{d y}{d x} = x

\int 1 + 2 x^{2} d x

f (x) = 3 cos (3 x)

\frac{d}{d x} (- \frac{6 x}{6 x ^{2} + 8})

\int_{- 3}^{2} π (y^{2} + 2 y + 4)^{2} d y