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

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

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

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

2 x - 3 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 2 x - 3 y + 0

Simplify

= 2 x - 3 y

\int 3 sin (ln (x)) d x

\int_{0}^{x} e^{t} d t

y^{- 1} + y^{- 2}

\int_{- 1}^{0} x^{2} - 1 d x

x \to - 9 + lim (\frac{1}{x + 9})

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