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

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

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

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

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derivative of

f (x) = \frac{( x + 1 )}{x - 1}

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

Solution steps

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial y} (x + 1)

\frac{d}{d x} (e^{g (x)} cos (x))

f (x) = - 2 x sin (x^{2})

x \to - \infty lim (- 2 x^{3})

x^{2} y^{^{''}} + 17 x y^{^{'}} + 145 y = 0

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