What is the derivative of y=(4x^3+2x+5)/x ?

The derivative of y=(4x^3+2x+5)/x is (8x^3-5)/(x^2)

What is the first derivative of y=(4x^3+2x+5)/x ?

The first derivative of y=(4x^3+2x+5)/x is (8x^3-5)/(x^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = \frac{4 x ^{3} + 2 x + 5}{x}

\frac{8 x ^{3} - 5}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 x ^{3} + 2 x + 5}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 4 x ^{3} + 2 x + 5 ) x - \frac{d x}{d x} ( 4 x ^{3} + 2 x + 5 )}{x ^{2}}

\frac{d}{d x} (4 x^{3} + 2 x + 5) = 12 x^{2} + 2

\frac{d x}{d x} = 1

= \frac{( 12 x ^{2} + 2 ) x - 1 \cdot ( 4 x ^{3} + 2 x + 5 )}{x ^{2}}

Simplify

\frac{( 12 x ^{2} + 2 ) x - 1 \cdot ( 4 x ^{3} + 2 x + 5 )}{x ^{2}} : \frac{8 x ^{3} - 5}{x ^{2}}

= \frac{8 x ^{3} - 5}{x ^{2}}

\int_{0}^{e} x \cdot ln (x) d x

\int 0.2 x + 8 d x

e^{x}, e^{- x}, x = - 2, x = 1

\int e^{- 0.125 x} d x

\frac{\partial}{\partial y} (e^{x y^{2}} + 4 z y^{3})

What is the derivative of y=(4x^3+2x+5)/x ?
The derivative of y=(4x^3+2x+5)/x is (8x^3-5)/(x^2)
What is the first derivative of y=(4x^3+2x+5)/x ?
The first derivative of y=(4x^3+2x+5)/x is (8x^3-5)/(x^2)