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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 x ( 2 + x ) - \frac{1}{2 x} x ^{2}}{( 2 + x ) ^{2}} : \frac{4 x ( 2 + x ) - x ^{\frac{3}{2}}}{2 ( 2 + x ) ^{2}}

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

\int x^{4} ln (9 x) d x

y^{^{''}} + 4 y^{^{'}} - 3 y = x

\frac{d}{d x} (sin^{2} (x) + cos^{2} (x))

\int e^{- \frac{t}{2}} d t

y^{^{''}} = 8 t^{2}

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