What is (\partial)/(\partial z)(ln(1+z^2)) ?

The answer to (\partial)/(\partial z)(ln(1+z^2)) is (2z)/(1+z^2)

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(\partial)/(\partial z)(ln(1+z^2))

\frac{\partial}{\partial z} (ln (1 + z^{2}))

\frac{2 z}{1 + z ^{2}}

Solution steps

\frac{\partial}{\partial z} (ln (1 + z^{2}))

Apply the chain rule:

\frac{1}{1 + z ^{2}} \frac{\partial}{\partial z} (1 + z^{2})

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

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

= \frac{1}{1 + z ^{2}} \cdot 2 z

Simplify

\frac{1}{1 + z ^{2}} \cdot 2 z : \frac{2 z}{1 + z ^{2}}

= \frac{2 z}{1 + z ^{2}}

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What is (\partial)/(\partial z)(ln(1+z^2)) ?
The answer to (\partial)/(\partial z)(ln(1+z^2)) is (2z)/(1+z^2)