What is the derivative of (6x^2+6x+8)/(sqrt(x)) ?

The derivative of (6x^2+6x+8)/(sqrt(x)) is (9x^2+3x-4)/(xsqrt(x))

What is the first derivative of (6x^2+6x+8)/(sqrt(x)) ?

The first derivative of (6x^2+6x+8)/(sqrt(x)) is (9x^2+3x-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{6 x ^{2} + 6 x + 8}{x}

\frac{9 x ^{2} + 3 x - 4}{x x}

Solution steps

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

\frac{d}{d x} (6 x^{2} + 6 x + 8) = 12 x + 6

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

= \frac{( 12 x + 6 ) x - \frac{1}{2 x} ( 6 x ^{2} + 6 x + 8 )}{( x ) ^{2}}

Simplify

\frac{( 12 x + 6 ) x - \frac{1}{2 x} ( 6 x ^{2} + 6 x + 8 )}{( x ) ^{2}} : \frac{9 x ^{2} + 3 x - 4}{x x}

= \frac{9 x ^{2} + 3 x - 4}{x x}

\int csc^{4} (9 x) d x

(x - 1) e^{x}

\frac{\partial}{\partial x} (7 x y + x cos (x))

- x^{2} + 4, 3 x

\int (8 x + 5)^{- 3} d x

What is the derivative of (6x^2+6x+8)/(sqrt(x)) ?
The derivative of (6x^2+6x+8)/(sqrt(x)) is (9x^2+3x-4)/(xsqrt(x))
What is the first derivative of (6x^2+6x+8)/(sqrt(x)) ?
The first derivative of (6x^2+6x+8)/(sqrt(x)) is (9x^2+3x-4)/(xsqrt(x))