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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

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

\int (4 x) d x

\frac{d y}{d t} = \frac{2 t}{1 + t ^{2}} y + \frac{2}{1 + t ^{2}}

\frac{d}{d x} (9 - 2 x)

f (x) = x^{2} - 6 x + 1, at x = 1

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