What is the derivative of y=(sqrt(x))/(4+x) ?

The derivative of y=(sqrt(x))/(4+x) is (4-x)/(2sqrt(x)(4+x)^2)

What is the first derivative of y=(sqrt(x))/(4+x) ?

The first derivative of y=(sqrt(x))/(4+x) is (4-x)/(2sqrt(x)(4+x)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = \frac{x}{4 + x}

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x) = \frac{1}{2 x}

\frac{d}{d x} (4 + x) = 1

= \frac{\frac{1}{2 x} ( 4 + x ) - 1 \cdot x}{( 4 + x ) ^{2}}

Simplify

\frac{\frac{1}{2 x} ( 4 + x ) - 1 \cdot x}{( 4 + x ) ^{2}} : \frac{4 - x}{2 x ( 4 + x ) ^{2}}

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

n = 1 \sum \infty \frac{4}{n ( n + 3 ) !}

\int 3 x^{2} y d x

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} + 4})

\frac{\partial}{\partial x} (2 x^{4} y + sin (x^{4}))

What is the derivative of y=(sqrt(x))/(4+x) ?
The derivative of y=(sqrt(x))/(4+x) is (4-x)/(2sqrt(x)(4+x)^2)
What is the first derivative of y=(sqrt(x))/(4+x) ?
The first derivative of y=(sqrt(x))/(4+x) is (4-x)/(2sqrt(x)(4+x)^2)