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

The first (1/(4x))^' is -1/(4x^2)

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(1/(4x))^'

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

- \frac{1}{4 x ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

= \frac{1}{4} (- 1 \cdot x^{- 1 - 1})

Simplify

\frac{1}{4} (- 1 \cdot x^{- 1 - 1}) : - \frac{1}{4 x ^{2}}

= - \frac{1}{4 x ^{2}}

\frac{d}{d x} (4 x + 8 x^{\frac{3}{4}})

x \to 2 lim ([f (x) + 5 g (x)])

\int csc (x) \cdot cot (x) d x

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

\frac{d}{d x} ((2 x^{2} + 3 x)^{2})