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

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

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

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

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derivative of

y = \frac{e ^{x}}{x - 1}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d y}{d x} = \frac{( e ^{- y} + e ^{- 5 x - y} )}{e ^{x} y}

x \to 1 lim (- 9 + 3 x^{2})

\int t^{2} - t^{4} \frac{d}{1} d t

y^{^{'''}} + 2 y^{^{''}} + 2 y^{^{'}} = 0

arctan (\frac{y}{x})

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