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

The derivative of (2x+1/(4x-1)) is -6/((4x-1)^2)

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

The first derivative of (2x+1/(4x-1)) is -6/((4x-1)^2)

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

- \frac{6}{( 4 x - 1 ) ^{2}}

Solution steps

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

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

\frac{d}{d x} (4 x - 1) = 4

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

Simplify

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

= - \frac{6}{( 4 x - 1 ) ^{2}}

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

\int sin (6)

\frac{\partial}{\partial u} (sin (u) cos (v))

\frac{d y}{d x} = e^{y} sin (x)

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

What is the derivative of (2x+1/(4x-1)) ?
The derivative of (2x+1/(4x-1)) is -6/((4x-1)^2)
What is the first derivative of (2x+1/(4x-1)) ?
The first derivative of (2x+1/(4x-1)) is -6/((4x-1)^2)