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

The derivative of (2x^2+8x+2/(sqrt(x))) is (3x^2+4x-1)/(xsqrt(x))

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

The first derivative of (2x^2+8x+2/(sqrt(x))) is (3x^2+4x-1)/(xsqrt(x))

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{2 x ^{2} + 8 x + 2}{x})

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

Solution steps

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

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

Simplify

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

= \frac{3 x ^{2} + 4 x - 1}{x x}

\frac{d y}{d x} = 2 - 2 x - y + 3

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

\int_{1}^{0} (- 9 x) d x

\frac{\partial}{\partial t} (ln (x + 2 c t))

\int_{- 1}^{2} ∣ 2 x - 3 ∣ d x

What is the derivative of (2x^2+8x+2/(sqrt(x))) ?
The derivative of (2x^2+8x+2/(sqrt(x))) is (3x^2+4x-1)/(xsqrt(x))
What is the first derivative of (2x^2+8x+2/(sqrt(x))) ?
The first derivative of (2x^2+8x+2/(sqrt(x))) is (3x^2+4x-1)/(xsqrt(x))