What is the first (4/(x^2))^' ?

The first (4/(x^2))^' is -8/(x^3)

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(4/(x^2))^'

(\frac{4}{x ^{2}})^{^{'}}

- \frac{8}{x ^{3}}

Solution steps

(\frac{4}{x ^{2}})^{'}

Take the constant out:

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

= 4 (\frac{1}{x ^{2}})^{^{'}}

Apply exponent rule:

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

= 4 (x^{- 2})^{^{'}}

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 (- 2 x^{- 2 - 1})

Simplify

4 (- 2 x^{- 2 - 1}) : - \frac{8}{x ^{3}}

= - \frac{8}{x ^{3}}

\frac{d}{d x} ((3 x^{4} - x^{3}) (x^{2} + x^{3}))

\frac{\partial}{\partial x} (x y^{3} + 6)

x^{2} y^{^{'}} + y = 0, y (8) = 5

\int_{0}^{3} \frac{1}{( x + 2 ) ( x + 4 )} d x

\frac{10 s}{s ^{2} - 25}