What is (\partial)/(\partial y)(2xyln(y)) ?

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

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(\partial)/(\partial y)(2xyln(y))

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

2 x (ln (y) + 1)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

= 2 x \frac{\partial}{\partial y} (y ln (y))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = ln (y)

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

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

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

= 2 x (1 \cdot ln (y) + \frac{1}{y} y)

Simplify

2 x (1 \cdot ln (y) + \frac{1}{y} y) : 2 x (ln (y) + 1)

= 2 x (ln (y) + 1)

\frac{5}{12} (3 + 5) x^{3}

\frac{\partial}{\partial v} (3 v cos (u))

\frac{( 1 + x )}{x}

\int \frac{y + 3}{( 3 - y ) ^{\frac{2}{3}}} d y

y = \frac{1}{x ^{2} + 1}, x = 0, x = 6, y = 0

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