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

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

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

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

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derivative of

y = \frac{2 x + 1}{2 x - 1}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

3 x - 45 - 15 y = 0

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

f (x) = \frac{1 - sec ( x )}{tan ( x )}

x = - 2

y = 2 x - 5

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