What is the derivative of f(x)=(x^2-2)/(2x+1) ?

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

2 x (2 x + 1) - 2 (x^{2} - 2) : 2 x^{2} + 2 x + 4

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

\int \frac{e ^{2 t}}{1 - e ^{2 t}} d t

\frac{d y}{d x} = x^{8} y^{- 5}

(3 x y^{2} - 3) d x + (3 x^{2} y + 4) d y = 0

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

\frac{d P}{d t} = 2 Pt, P (1) = 7

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