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

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

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(-1/(x^2))^'

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

\frac{2}{x ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

= \frac{2}{x ^{3}}

\frac{d}{d x} (x ln (x^{2}) - 2 x)

\int e^{2 x} \frac{1}{2} d x

\int csc (2 y) d y

\frac{\partial}{\partial x} (y^{2} - x^{2})

ln (cos (x))