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

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

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

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

- x

Solution steps

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

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

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

= 0 - x

Simplify

= - x

\frac{2 x}{x + 1}

\int \frac{1}{e ^{- 2 y}} d y

\int_{0}^{\frac{1}{2}} 30 x d x

\frac{d}{d x} (\frac{x + 1}{x ^{2} + 1})

\int \frac{1}{4} x^{4} d x

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