What is the derivative of f(z)=sqrt((z-4)/(z+4)) ?

The derivative of f(z)=sqrt((z-4)/(z+4)) is 4/(sqrt(z-4)(z+4)^{3/2)}

What is the first derivative of f(z)=sqrt((z-4)/(z+4)) ?

The first derivative of f(z)=sqrt((z-4)/(z+4)) is 4/(sqrt(z-4)(z+4)^{3/2)}

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derivative of

f (z) = \frac{z - 4}{z + 4}

\frac{4}{z - 4 ( z + 4 ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d z} (\frac{z - 4}{z + 4})

Apply the chain rule:

\frac{1}{2 \frac{z - 4}{z + 4}} \frac{d}{d z} (\frac{z - 4}{z + 4})

= \frac{1}{2 \frac{z - 4}{z + 4}} \frac{d}{d z} (\frac{z - 4}{z + 4})

\frac{d}{d z} (\frac{z - 4}{z + 4}) = \frac{8}{( z + 4 ) ^{2}}

= \frac{1}{2 \frac{z - 4}{z + 4}} \cdot \frac{8}{( z + 4 ) ^{2}}

Simplify

\frac{1}{2 \frac{z - 4}{z + 4}} \cdot \frac{8}{( z + 4 ) ^{2}} : \frac{4}{z - 4 ( z + 4 ) ^{\frac{3}{2}}}

= \frac{4}{z - 4 ( z + 4 ) ^{\frac{3}{2}}}

\frac{\partial}{\partial x} (x ln (z^{2} + x^{7}))

\frac{d}{d x} (x^{4} e^{- x})

\frac{d}{d x} ((csc (x) + cot (x))^{- 1})

n = 0 \sum \infty - \frac{n}{2}

x^{4} + 6 x^{2} - x

What is the derivative of f(z)=sqrt((z-4)/(z+4)) ?
The derivative of f(z)=sqrt((z-4)/(z+4)) is 4/(sqrt(z-4)(z+4)^{3/2)}
What is the first derivative of f(z)=sqrt((z-4)/(z+4)) ?
The first derivative of f(z)=sqrt((z-4)/(z+4)) is 4/(sqrt(z-4)(z+4)^{3/2)}