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}})^{'}

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}}

\int sin (x) sec^{7} (x) d x

\frac{\partial}{\partial z} (e^{- x yz})

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

\frac{d}{d x} ((- 2 x - 3 x^{5}) tan (x))

h \to 0 lim (\frac{7 x h + h ^{2}}{h})