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}} \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 \frac{6 x - 1}{4 x ^{2} + 4 x + 10} d x

\int \frac{28}{( 196 x ^{2} + 1 ) ^{2}} d x

y^{'} = y (5 - y)

\frac{\partial}{\partial y} (- 4 e^{x} cos (yz))

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

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)}