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

The derivative of y=(6x^2+2x+2)/(sqrt(x)) is (9x^2+x-1)/(xsqrt(x))

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

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

\frac{9 x ^{2} + x - 1}{x x}

Solution steps

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

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

Simplify

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

= \frac{9 x ^{2} + x - 1}{x x}

\frac{d}{d x} (ln (4 x^{3} - x^{2}))

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

y^{^{''}} + 4 y^{^{'}} + 24 y = 0

x \to 1 lim (x^{2} + 5 x - 6)

\frac{\partial}{\partial y} (x sin (y))

What is the derivative of y=(6x^2+2x+2)/(sqrt(x)) ?
The derivative of y=(6x^2+2x+2)/(sqrt(x)) is (9x^2+x-1)/(xsqrt(x))
What is the first derivative of y=(6x^2+2x+2)/(sqrt(x)) ?
The first derivative of y=(6x^2+2x+2)/(sqrt(x)) is (9x^2+x-1)/(xsqrt(x))