What is the (1/(2sqrt(x)))^' ?

The (1/(2sqrt(x)))^' is -1/(4x^{3/2)}

What is the first (1/(2sqrt(x)))^' ?

The first (1/(2sqrt(x)))^' is -1/(4x^{3/2)}

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(\frac{1}{2 x})^{^{'}}

- \frac{1}{4 x ^{\frac{3}{2}}}

Solution steps

(\frac{1}{2 x})^{'}

Take the constant out:

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

= \frac{1}{2} (\frac{1}{x})^{^{'}}

Apply exponent rule:

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

= \frac{1}{2} (x^{- \frac{1}{2}})^{^{'}}

Apply the Power Rule:

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

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

Simplify

\frac{1}{2} (- \frac{1}{2} x^{- \frac{1}{2} - 1}) : - \frac{1}{4 x ^{\frac{3}{2}}}

= - \frac{1}{4 x ^{\frac{3}{2}}}

θ cos (θ)

\int (2 x - 4^{x}) d x

y = x^{2} - 4, y = 5

x \to \infty lim (1^{- x})

\frac{d}{d x} (8^{s i n (π x)})

What is the (1/(2sqrt(x)))^' ?
The (1/(2sqrt(x)))^' is -1/(4x^{3/2)}
What is the first (1/(2sqrt(x)))^' ?
The first (1/(2sqrt(x)))^' is -1/(4x^{3/2)}