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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (4 x^{2} - e^{x})

\frac{d y}{d x} = - \frac{1}{5 x y}

\int (\frac{ln ( x )}{x})^{2} d x

\frac{\partial}{\partial x} (300 - 20 x^{2} + 30 y)

x \frac{d y}{d x} + 1 y = 6 y^{2}, y (1) = - 1

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