What is the derivative of 9/(4\sqrt[4]{x^3}) ?

The derivative of 9/(4\sqrt[4]{x^3}) is -(27)/(16x^{7/4)}

What is the first derivative of 9/(4\sqrt[4]{x^3}) ?

The first derivative of 9/(4\sqrt[4]{x^3}) is -(27)/(16x^{7/4)}

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\frac{d}{d x} (\frac{9}{4 4 x ^{3}})

- \frac{27}{16 x ^{\frac{7}{4}}}

Solution steps

\frac{d}{d x} (\frac{9}{4 4 x ^{3}})

Take the constant out:

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

= \frac{9}{4} \frac{d}{d x} (\frac{1}{4 x ^{3}})

Apply exponent rule:

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

= \frac{9}{4} \frac{d}{d x} ((x^{3})^{- \frac{1}{4}})

Apply the chain rule:

- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \frac{d}{d x} (x^{3})

= - \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \frac{d}{d x} (x^{3})

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

= \frac{9}{4} (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2})

Simplify

\frac{9}{4} (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2}) : - \frac{27}{16 x ^{\frac{7}{4}}}

= - \frac{27}{16 x ^{\frac{7}{4}}}

\int \frac{2 x ^{2} - x}{x ^{2} - x + 2} d x

\int 2 x e^{x} d x

\int (2 x - 1) y^{2} sin (4 x y) d x

d er i v a t i v e f (x) = 8 x - 3

t an g e n t f (x) = 2 e^{x} cos (x), at x = 0

What is the derivative of 9/(4\sqrt[4]{x^3}) ?
The derivative of 9/(4\sqrt[4]{x^3}) is -(27)/(16x^{7/4)}
What is the first derivative of 9/(4\sqrt[4]{x^3}) ?
The first derivative of 9/(4\sqrt[4]{x^3}) is -(27)/(16x^{7/4)}