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

The answer to (\partial)/(\partial y)(x^2+2y^2-x) is 4y

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(\partial)/(\partial y)(x^2+2y^2-x)

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

4 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (2 y^{2}) = 4 y

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

= 0 + 4 y - 0

Simplify

= 4 y

\int \frac{2 x - 3}{4 x - x ^{2}} d x

8

\int e^{t a n (4 x)} sec^{2} (4 x) d x

\frac{1}{1 - x}

\int_{e}^{e^{2}} \frac{1}{x ln ( x )} d x

What is (\partial)/(\partial y)(x^2+2y^2-x) ?
The answer to (\partial)/(\partial y)(x^2+2y^2-x) is 4y