What is (\partial)/(\partial y)(ln(3xy)+x^2) ?

The answer to (\partial)/(\partial y)(ln(3xy)+x^2) is 1/y

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(\partial)/(\partial y)(ln(3xy)+x^2)

\frac{\partial}{\partial y} (ln (3 x y) + x^{2})

\frac{1}{y}

Solution steps

\frac{\partial}{\partial y} (ln (3 x y) + x^{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (ln (3 x y)) + \frac{\partial}{\partial y} (x^{2})

\frac{\partial}{\partial y} (ln (3 x y)) = \frac{1}{y}

\frac{\partial}{\partial y} (x^{2}) = 0

= \frac{1}{y} + 0

Simplify

= \frac{1}{y}

\frac{\partial}{\partial y} (\frac{y}{( 2 x + 3 y )})

x \to \infty lim (- 1)

x \to 4 lim (2 x)

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

\frac{\partial}{\partial y} (1 + 3 x + 2 y^{2})

What is (\partial)/(\partial y)(ln(3xy)+x^2) ?
The answer to (\partial)/(\partial y)(ln(3xy)+x^2) is 1/y