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

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

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(1/(x^3))^'

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

- \frac{3}{x ^{4}}

Solution steps

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

Apply exponent rule:

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

= (x^{- 3})^{^{'}}

Apply the Power Rule:

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

= - 3 x^{- 3 - 1}

Simplify

- 3 x^{- 3 - 1} : - \frac{3}{x ^{4}}

= - \frac{3}{x ^{4}}

y = (3 x)^{5 x}

y = 9 x, y = \frac{3}{4} x, y = 70 - x^{2}

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

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