What is the derivative of (4x^3-2x+1/(x^2)) ?

The derivative of (4x^3-2x+1/(x^2)) is (2(2x^3+x-1))/(x^3)

What is the first derivative of (4x^3-2x+1/(x^2)) ?

The first derivative of (4x^3-2x+1/(x^2)) is (2(2x^3+x-1))/(x^3)

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 ^{3} - 2 x + 1}{x ^{2}})

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

Solution steps

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

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

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

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

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

Simplify

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

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

\int_{0}^{2 π} 3 sin (2 x) d x

x \to 0 lim (\frac{- ln ( 1 + 6 x )}{x})

\int \frac{x ^{3}}{1 + x ^{2} - x ^{4}} d x

\frac{d}{d x} (x^{3} sin (\frac{1}{x}))

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

What is the derivative of (4x^3-2x+1/(x^2)) ?
The derivative of (4x^3-2x+1/(x^2)) is (2(2x^3+x-1))/(x^3)
What is the first derivative of (4x^3-2x+1/(x^2)) ?
The first derivative of (4x^3-2x+1/(x^2)) is (2(2x^3+x-1))/(x^3)