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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

y^{^{'}} - y = e^{2 x}

\int 12 cos (4 x) d x

f (x) = 5 x + 1

x \to 0 - lim (\frac{x ^{2}}{6} - \frac{5}{x})

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

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