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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

x \to 0 lim ((\frac{2 arccos ( x )}{π})^{\frac{1}{x}})

\int \frac{2 + x + x ^{2}}{x} d x

f (t) = \frac{sin ( t )}{t}

\int \frac{y}{e ^{- y}} d y

f (x) = \frac{x cos ( x ) - 2 π}{x - 2 π}

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