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

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

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(\partial)/(\partial x)(-ln(x^2+1))

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

- \frac{2 x}{x ^{2} + 1}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{\partial}{\partial x} (ln (x^{2} + 1))

Apply the chain rule:

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

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

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

= - \frac{1}{x ^{2} + 1} \cdot 2 x

Simplify

- \frac{1}{x ^{2} + 1} \cdot 2 x : - \frac{2 x}{x ^{2} + 1}

= - \frac{2 x}{x ^{2} + 1}

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