What is the derivative of (x^2/(x^3+4)) ?

The derivative of (x^2/(x^3+4)) is (-x^4+8x)/((x^3+4)^2)

What is the first derivative of (x^2/(x^3+4)) ?

The first derivative of (x^2/(x^3+4)) is (-x^4+8x)/((x^3+4)^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 ^{2}}{x ^{3} + 4})

\frac{- x ^{4} + 8 x}{( x ^{3} + 4 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 x ( x ^{3} + 4 ) - 3 x ^{2} x ^{2}}{( x ^{3} + 4 ) ^{2}} : \frac{- x ^{4} + 8 x}{( x ^{3} + 4 ) ^{2}}

= \frac{- x ^{4} + 8 x}{( x ^{3} + 4 ) ^{2}}

\frac{\partial}{\partial x} (5 y - x + y^{2} sin (y))

\frac{d y}{d x} = y^{3} x e^{x^{2}}

y^{^{''}} - 8 y = 0

e^{x + y}

\int_{1}^{4} 5 x + 3 x^{2} - 8 d x

What is the derivative of (x^2/(x^3+4)) ?
The derivative of (x^2/(x^3+4)) is (-x^4+8x)/((x^3+4)^2)
What is the first derivative of (x^2/(x^3+4)) ?
The first derivative of (x^2/(x^3+4)) is (-x^4+8x)/((x^3+4)^2)