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

The derivative of y=(sqrt(x)+1)/(sqrt(x)-1) is -1/(sqrt(x)(\sqrt{x)-1)^2}

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

The first derivative of y=(sqrt(x)+1)/(sqrt(x)-1) is -1/(sqrt(x)(\sqrt{x)-1)^2}

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derivative of

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

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

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

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

Simplify

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

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

n = 2 \sum \infty \frac{1}{n ln ( n )}

x \to 0 lim (\frac{4 cos ( x ) - 4}{x})

\frac{\partial}{\partial x} (6 x yz)

\frac{d}{d x} (-, x)

(- 8, - 9), (- 5, - 3)

What is the derivative of y=(sqrt(x)+1)/(sqrt(x)-1) ?
The derivative of y=(sqrt(x)+1)/(sqrt(x)-1) is -1/(sqrt(x)(\sqrt{x)-1)^2}
What is the first derivative of y=(sqrt(x)+1)/(sqrt(x)-1) ?
The first derivative of y=(sqrt(x)+1)/(sqrt(x)-1) is -1/(sqrt(x)(\sqrt{x)-1)^2}