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

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

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

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

\frac{1}{x} + y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

= \frac{1}{x} + y

\int (sech (x)) d x

8 e^{π - t}

\int sin (8 x) cos (12 x) d x

ln (x), 4

g (x) = 1 - 2 x - x^{(2)}

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