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

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

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(\partial)/(\partial y)(xy+ln(x+y))

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

y = 2 x e^{x}, (0, 0)

cos (π x)

\frac{d y}{d x} + y cos (x) = 3 cos (x), y (0) = 5

y = 4 x - x^{2}, y = x

f (x) = \frac{5}{x}, at x = \frac{1}{16}

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