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

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

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(1/(3x))^'

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

- \frac{1}{3 x ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

= \frac{1}{3} (- 1 \cdot x^{- 1 - 1})

Simplify

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

= - \frac{1}{3 x ^{2}}

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

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