What is the derivative of (2x^2+8x+8)/(sqrt(x)) ?

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

What is the first derivative of (2x^2+8x+8)/(sqrt(x)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{2 x ^{2} + 8 x + 8}{x}

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

Solution steps

\frac{d}{d x} (\frac{2 x ^{2} + 8 x + 8}{x})

Apply the Quotient Rule:

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

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

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

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

= \frac{( 4 x + 8 ) x - \frac{1}{2 x} ( 2 x ^{2} + 8 x + 8 )}{( x ) ^{2}}

Simplify

\frac{( 4 x + 8 ) x - \frac{1}{2 x} ( 2 x ^{2} + 8 x + 8 )}{( x ) ^{2}} : \frac{3 x ^{2} + 4 x - 4}{x x}

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

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

\int_{2}^{4} 3 x - 6 d x

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

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

x \to 4 - lim (\frac{4}{x - 4})

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