What is the derivative of y= 4/(sqrt(x)) ?

The derivative of y= 4/(sqrt(x)) is -2/(x^{3/2)}

What is the first derivative of y= 4/(sqrt(x)) ?

The first derivative of y= 4/(sqrt(x)) is -2/(x^{3/2)}

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derivative of

y = \frac{4}{x}

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

Solution steps

\frac{d}{d x} (\frac{4}{x})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{1}{x})

Apply exponent rule:

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

= 4 \frac{d}{d x} (x^{- \frac{1}{2}})

Apply the Power Rule:

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

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

Simplify

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

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

\int 12 (x - 4)^{5} d x

9 x - 36

\frac{\partial}{\partial x} (- 4 x + 5 - 6 x y)

f (x) = \frac{x ^{3}}{3}

What is the derivative of y= 4/(sqrt(x)) ?
The derivative of y= 4/(sqrt(x)) is -2/(x^{3/2)}
What is the first derivative of y= 4/(sqrt(x)) ?
The first derivative of y= 4/(sqrt(x)) is -2/(x^{3/2)}