What is the first (1/(x^4))^' ?

The first (1/(x^4))^' is -4/(x^5)

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(1/(x^4))^'

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

- \frac{4}{x ^{5}}

Solution steps

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

= - 4 x^{- 4 - 1}

Simplify

- 4 x^{- 4 - 1} : - \frac{4}{x ^{5}}

= - \frac{4}{x ^{5}}

f (x) = arccos (e^{2 x})

\frac{\partial}{\partial x} (4 - x^{4} + 2 x^{2} - y^{2})

\frac{\partial}{\partial x} (4 x - 2 y^{2} + z^{2})

x \to 6 lim (\frac{x - 3}{x - 1})