What is the derivative of (x^4/(4-x^3)) ?

The derivative of (x^4/(4-x^3)) is (16x^3-x^6)/((4-x^3)^2)

What is the first derivative of (x^4/(4-x^3)) ?

The first derivative of (x^4/(4-x^3)) is (16x^3-x^6)/((4-x^3)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\frac{16 x ^{3} - x ^{6}}{( 4 - x ^{3} ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x ^{4} ) ( 4 - x ^{3} ) - \frac{d}{d x} ( 4 - x ^{3} ) x ^{4}}{( 4 - x ^{3} ) ^{2}}

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

\frac{d}{d x} (4 - x^{3}) = - 3 x^{2}

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

Simplify

\frac{4 x ^{3} ( 4 - x ^{3} ) - ( - 3 x ^{2} ) x ^{4}}{( 4 - x ^{3} ) ^{2}} : \frac{16 x ^{3} - x ^{6}}{( 4 - x ^{3} ) ^{2}}

= \frac{16 x ^{3} - x ^{6}}{( 4 - x ^{3} ) ^{2}}

cos (x) - 4 tan (x)

\frac{\partial}{\partial x} (8 - x^{2} - 3 y^{2})

(2 x^{3} + 4) (x^{4} - 4 x)

\frac{6}{x ^{3}} - \frac{6}{x ^{2}} - \frac{1}{x} + 350

\frac{d}{d x} (x^{3} + 5 x)

What is the derivative of (x^4/(4-x^3)) ?
The derivative of (x^4/(4-x^3)) is (16x^3-x^6)/((4-x^3)^2)
What is the first derivative of (x^4/(4-x^3)) ?
The first derivative of (x^4/(4-x^3)) is (16x^3-x^6)/((4-x^3)^2)