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

The answer to (\partial)/(\partial y)(ln((xy)/(y^2+z^2))) is (-y^2+z^2)/(y(y^2+z^2))

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(\partial)/(\partial y)(ln((xy)/(y^2+z^2)))

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

\frac{- y ^{2} + z ^{2}}{y ( y ^{2} + z ^{2} )}

Solution steps

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

Treat

x, z

as constants

Apply the chain rule:

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

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

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

= \frac{1}{\frac{x y}{y ^{2} + z ^{2}}} \cdot \frac{x ( - y ^{2} + z ^{2} )}{( y ^{2} + z ^{2} ) ^{2}}

Simplify

\frac{1}{\frac{x y}{y ^{2} + z ^{2}}} \cdot \frac{x ( - y ^{2} + z ^{2} )}{( y ^{2} + z ^{2} ) ^{2}} : \frac{- y ^{2} + z ^{2}}{y ( y ^{2} + z ^{2} )}

= \frac{- y ^{2} + z ^{2}}{y ( y ^{2} + z ^{2} )}

\frac{\partial}{\partial y} (y e^{y x})

3 cos (x) - 2 sin (x)

\int 3 x^{2} + 2 d x

x \to 0 + lim (x - 1)

y = 5 x, x = 5

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