What is (\partial)/(\partial x)(x^2+y^2+48-xy-6x) ?

The answer to (\partial)/(\partial x)(x^2+y^2+48-xy-6x) is 2x-y-6

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(\partial)/(\partial x)(x^2+y^2+48-xy-6x)

\frac{\partial}{\partial x} (x^{2} + y^{2} + 48 - x y - 6 x)

2 x - y - 6

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} + 48 - x y - 6 x)

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} (y^{2}) + \frac{\partial}{\partial x} (48) - \frac{\partial}{\partial x} (x y) - \frac{\partial}{\partial x} (6 x)

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

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

\frac{\partial}{\partial x} (48) = 0

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

\frac{\partial}{\partial x} (6 x) = 6

= 2 x + 0 + 0 - y - 6

Simplify

= 2 x - y - 6

\frac{cos ( x )}{x ^{7}}

f (x) = x^{2} - 3 x

\int 21 t e^{t} d t

\int 3 x x^{2} + 4 d x

\frac{d}{d x} (e^{- 5 x} cos (3 x))

What is (\partial)/(\partial x)(x^2+y^2+48-xy-6x) ?
The answer to (\partial)/(\partial x)(x^2+y^2+48-xy-6x) is 2x-y-6