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

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

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

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

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derivative of

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

Simplify

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

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

\int 7 e^{2 x + e^{2 x}} d x

\frac{s ^{2}}{s ^{2} + 9}

\int \frac{2 e ^{2 x}}{e ^{2 x} + 16 e ^{x} + 63} d x

ln (x^{6})

\int_{0}^{1} \frac{x ^{4}}{x ^{5} - 1} d x

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