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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

Solution steps

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

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

Simplify

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

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

y = 3 x^{3} - x^{2} + 8, (2, 28)

\int (t^{2} - 2 cos (t)) d t

f (x) = 2 x^{2} - 3 x + 1

x \to 0 lim (\frac{100}{x ^{2}})

cos^{3} (t)

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