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

The derivative of sqrt((z+1)/(z-1)) is -1/(sqrt(z+1)(z-1)^{3/2)}

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

The first derivative of sqrt((z+1)/(z-1)) is -1/(sqrt(z+1)(z-1)^{3/2)}

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derivative of

\frac{z + 1}{z - 1}

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

Solution steps

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

Apply the chain rule:

\frac{1}{2 \frac{z + 1}{z - 1}} \frac{d}{d z} (\frac{z + 1}{z - 1})

= \frac{1}{2 \frac{z + 1}{z - 1}} \frac{d}{d z} (\frac{z + 1}{z - 1})

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

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

Simplify

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

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

\frac{1}{3 s + 4}

\int \frac{4}{x ^{2} x ^{2} + 36} d x

y^{^{''}} - 8 y^{^{'}} + 19 y = 0, y (0) = - 4, y^{^{'}} (0) = 2

\int_{1}^{2} \frac{x}{x ^{2} + 1} d x

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

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