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

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

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

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

2 x + y - 3

Solution steps

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

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

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

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

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

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

= 2 x + y + 0 - 3 + 0

Simplify

= 2 x + y - 3

x \to - 3 lim (- x^{2} + 6 x - 5)

f (x) = - 2 sin (x), at x = \frac{π}{4}

\int \frac{y ^{3}}{y ^{2} + 4} d y

f (x) = 4 tan (x), at x = \frac{π}{4}

\int_{- 2}^{3} \frac{43}{x ^{4}} d x

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