What is (\partial)/(\partial y)((x-4)ln(xy)) ?

The answer to (\partial)/(\partial y)((x-4)ln(xy)) is (x-4)/y

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(\partial)/(\partial y)((x-4)ln(xy))

\frac{\partial}{\partial y} ((x - 4) ln (x y))

\frac{x - 4}{y}

Solution steps

\frac{\partial}{\partial y} ((x - 4) ln (x y))

Treat

x

as a constant

Take the constant out:

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

= (x - 4) \frac{\partial}{\partial y} (ln (x y))

Apply the chain rule:

\frac{1}{x y} \frac{\partial}{\partial y} (x y)

= \frac{1}{x y} \frac{\partial}{\partial y} (x y)

\frac{\partial}{\partial y} (x y) = x

= (x - 4) \frac{1}{x y} x

Simplify

(x - 4) \frac{1}{x y} x : \frac{x - 4}{y}

= \frac{x - 4}{y}

θ \to 0 lim (sin (θ))

\frac{1}{2 r} + \frac{1}{8 r ^{\frac{7}{8}}}

\int_{0}^{l n (4)} \frac{e ^{x}}{e ^{2 x} + 25} d x

\frac{d}{d x} (ln (csc (x) - tan (x)))

x \to - \infty lim (\frac{1}{x} + 3)

What is (\partial)/(\partial y)((x-4)ln(xy)) ?
The answer to (\partial)/(\partial y)((x-4)ln(xy)) is (x-4)/y