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

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

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

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

\frac{1}{y} + x

Solution steps

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

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

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

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

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

= x + \frac{1}{y} + 0

Simplify

= \frac{1}{y} + x

\frac{\partial}{\partial x} (cos^{3} (x) y - 1)

\frac{s ^{2}}{( s ^{2} + 4 ) ^{2}}

(2 y + 3 x) d x = - x d y

\frac{d}{d x} (\frac{12}{x ^{2}})

y^{^{''}} + y^{^{'}} + 3 y = 3 t^{2}

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