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

The derivative of y=(4x^2+6x+2)/(sqrt(x)) is (6x^2+3x-1)/(xsqrt(x))

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

The first derivative of y=(4x^2+6x+2)/(sqrt(x)) is (6x^2+3x-1)/(xsqrt(x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

\frac{6 x ^{2} + 3 x - 1}{x x}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x^{2} + 6 x + 2) = 8 x + 6

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

= \frac{( 8 x + 6 ) x - \frac{1}{2 x} ( 4 x ^{2} + 6 x + 2 )}{( x ) ^{2}}

Simplify

\frac{( 8 x + 6 ) x - \frac{1}{2 x} ( 4 x ^{2} + 6 x + 2 )}{( x ) ^{2}} : \frac{6 x ^{2} + 3 x - 1}{x x}

= \frac{6 x ^{2} + 3 x - 1}{x x}

\int e^{2 x} \cdot e^{- x} d x

\int_{1}^{2} \frac{( x - 4 )}{x ^{2}} d x

f (x) = 2 x^{3} - 2 x, (1, 0)

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

\int 2 arctan (3 y) d y

What is the derivative of y=(4x^2+6x+2)/(sqrt(x)) ?
The derivative of y=(4x^2+6x+2)/(sqrt(x)) is (6x^2+3x-1)/(xsqrt(x))
What is the first derivative of y=(4x^2+6x+2)/(sqrt(x)) ?
The first derivative of y=(4x^2+6x+2)/(sqrt(x)) is (6x^2+3x-1)/(xsqrt(x))