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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

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

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

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

Simplify

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

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

n = 1 \sum \infty 6^{n} x^{n} n!

\int x cos (9 x^{2}) d x

y = (3 x^{3} - 2)^{5}

\frac{3 e ^{- 4 s}}{s + 2}

\int \frac{x ^{2} - 25}{x ^{2}} d x

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