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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial y} ((x^{2} + y^{2})^{\frac{1}{3}})

y^{^{'}} - y = 2 t e^{2 t}, y (0) = 1

sin (x y)

\int (\frac{5}{x} - 2 x^{3}) d x

\frac{d y}{d x} - e^{5 x + y} = 0

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