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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

\int \frac{- 3 x ^{3}}{x ^{2} + 9} d x

\frac{d}{d x} (\frac{x y}{z})

x \to 7 lim (2 cos (x))

\frac{d}{d x} ((1 + x)^{3 x})

x y^{^{'}} + y = 25 x

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