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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 4 lim (\frac{x - 4}{x ^{2} + 3 x - 28})

x \to \infty lim (\frac{x}{12})

\frac{d}{d x} (\frac{x - 7}{x + 5})

\frac{d}{d x} (\frac{0.3 ( 9 ^{x} )}{x ^{3}})

\frac{\partial}{\partial x} (x^{2} yz - x y z^{5})

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