What is the derivative of 1/(4x^{3/4)} ?

The derivative of 1/(4x^{3/4)} is -3/(16x^{7/4)}

What is the first derivative of 1/(4x^{3/4)} ?

The first derivative of 1/(4x^{3/4)} is -3/(16x^{7/4)}

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derivative of

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

- \frac{3}{16 x ^{\frac{7}{4}}}

Solution steps

\frac{d}{d x} (\frac{1}{4 x ^{\frac{3}{4}}})

Take the constant out:

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

= \frac{1}{4} \frac{d}{d x} (\frac{1}{x ^{\frac{3}{4}}})

Apply exponent rule:

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

= \frac{1}{4} \frac{d}{d x} (x^{- \frac{3}{4}})

Apply the Power Rule:

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

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

Simplify

\frac{1}{4} (- \frac{3}{4} x^{- \frac{3}{4} - 1}) : - \frac{3}{16 x ^{\frac{7}{4}}}

= - \frac{3}{16 x ^{\frac{7}{4}}}

(- 7.4) (- 4.1)

- sin (x)

(\frac{2}{2 s + 1}) - (\frac{1}{s})

(y^{2} + x y) d x - x^{2} d y = 0

\int (\frac{1}{t}) d t

What is the derivative of 1/(4x^{3/4)} ?
The derivative of 1/(4x^{3/4)} is -3/(16x^{7/4)}
What is the first derivative of 1/(4x^{3/4)} ?
The first derivative of 1/(4x^{3/4)} is -3/(16x^{7/4)}