What is the derivative of 9/(8\sqrt[4]{x^3}) ?

The derivative of 9/(8\sqrt[4]{x^3}) is -(27)/(32x^{7/4)}

What is the first derivative of 9/(8\sqrt[4]{x^3}) ?

The first derivative of 9/(8\sqrt[4]{x^3}) is -(27)/(32x^{7/4)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- \frac{27}{32 x ^{\frac{7}{4}}}

Solution steps

Take the constant out:

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

Apply exponent rule:

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

= \frac{9}{8} \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}{8} (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2})

Simplify

\frac{9}{8} (- \frac{1}{4 ( x ^{3} ) ^{\frac{5}{4}}} \cdot 3 x^{2}) : - \frac{27}{32 x ^{\frac{7}{4}}}

= - \frac{27}{32 x ^{\frac{7}{4}}}

\int_{0}^{1} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

\frac{d}{d x} ((8 x)^{\frac{1}{3}})

\int - cos (t) sin^{2} (t) d t

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + 5})

\int 3 cos^{2} (x) sin (2 x) d x

What is the derivative of 9/(8\sqrt[4]{x^3}) ?
The derivative of 9/(8\sqrt[4]{x^3}) is -(27)/(32x^{7/4)}
What is the first derivative of 9/(8\sqrt[4]{x^3}) ?
The first derivative of 9/(8\sqrt[4]{x^3}) is -(27)/(32x^{7/4)}