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

The answer to (\partial)/(\partial x)((3-x)(3-y)(x+y-3)) is (3-y)(6-2x-y)

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(\partial)/(\partial x)((3-x)(3-y)(x+y-3))

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

(3 - y) (6 - 2 x - y)

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 3 - x, g = x + y - 3

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

\frac{\partial}{\partial x} (3 - x) = - 1

\frac{\partial}{\partial x} (x + y - 3) = 1

= (3 - y) ((- 1) (x + y - 3) + 1 \cdot (3 - x))

Simplify

(3 - y) ((- 1) (x + y - 3) + 1 \cdot (3 - x)) : (3 - y) (6 - 2 x - y)

= (3 - y) (6 - 2 x - y)

\frac{d ^{2}}{d x ^{2}} ((3 + 2 x) e^{- 3 x})

x (- 3 x^{- 4})

\int - 5 sin^{3} (x) cos^{2} (x) d x

\frac{d y}{d t} = 1 0^{- 21} \cdot y^{3}

f (x) = - 3 x^{2} + 8 x + 1, at x = - 2

What is (\partial)/(\partial x)((3-x)(3-y)(x+y-3)) ?
The answer to (\partial)/(\partial x)((3-x)(3-y)(x+y-3)) is (3-y)(6-2x-y)