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

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

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

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

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

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

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 - 1}

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

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

Simplify

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

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

\frac{d}{d x} (x^{2} + x^{2} ln (x))

\frac{d y}{d t} = t^{4} y

x^{4}, 1

- \frac{2 x}{x ^{2} + 1}

\int_{0}^{l n (2)} s e^{s} d s

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