What is the derivative of (6x^2+2x+4)/(sqrt(x)) ?

The derivative of (6x^2+2x+4)/(sqrt(x)) is (9x^2+x-2)/(xsqrt(x))

What is the first derivative of (6x^2+2x+4)/(sqrt(x)) ?

The first derivative of (6x^2+2x+4)/(sqrt(x)) is (9x^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 >

derivative of

\frac{6 x ^{2} + 2 x + 4}{x}

\frac{9 x ^{2} + x - 2}{x x}

Solution steps

\frac{d}{d x} (\frac{6 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} ( 6 x ^{2} + 2 x + 4 ) x - \frac{d}{d x} ( x ) ( 6 x ^{2} + 2 x + 4 )}{( x ) ^{2}}

\frac{d}{d x} (6 x^{2} + 2 x + 4) = 12 x + 2

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

= \frac{( 12 x + 2 ) x - \frac{1}{2 x} ( 6 x ^{2} + 2 x + 4 )}{( x ) ^{2}}

Simplify

\frac{( 12 x + 2 ) x - \frac{1}{2 x} ( 6 x ^{2} + 2 x + 4 )}{( x ) ^{2}} : \frac{9 x ^{2} + x - 2}{x x}

= \frac{9 x ^{2} + x - 2}{x x}

y^{^{''}} + 3 y^{^{'}} - 2 y = 0

\frac{d y}{d x} = 4 e^{x - y}

(x y + y^{2} + y) d x + (x + 2 y) d y = 0

y^{^{'}} = cos (x + y)

\frac{\partial}{\partial x} (e^{x} cos (x + y))

What is the derivative of (6x^2+2x+4)/(sqrt(x)) ?
The derivative of (6x^2+2x+4)/(sqrt(x)) is (9x^2+x-2)/(xsqrt(x))
What is the first derivative of (6x^2+2x+4)/(sqrt(x)) ?
The first derivative of (6x^2+2x+4)/(sqrt(x)) is (9x^2+x-2)/(xsqrt(x))