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

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

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

The first derivative of (2x^2+6x+4)/(sqrt(x)) is (3x^2+3x-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{2 x ^{2} + 6 x + 4}{x}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{ln ( 3 x )}{7 x ^{2}})

f (x) = \frac{1}{x}

x \to - 8 lim (\frac{x - 8}{∣ x + 8 ∣})

\int \frac{1}{x ( ln ( x ) ) ^{π}} d x

n = 0 \sum \infty (- 1)^{n} 2^{n}

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