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

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

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

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

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

Solution steps

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

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

Simplify

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

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

\int sin (4 t) d t

\int sin (x) sin (2 x) d x

y^{^{'}} = 4 x + y, y (0) = 1

\int_{0}^{6} 6 x - (3 x - 12) d x

\int_{0}^{\infty} \frac{1}{7 + x} d x

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