What is (\partial)/(\partial x)(e^{x+y}(x-1)(y-2)) ?

The answer to (\partial)/(\partial x)(e^{x+y}(x-1)(y-2)) is e^{x+y}x(y-2)

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(\partial)/(\partial x)(e^{x+y}(x-1)(y-2))

\frac{\partial}{\partial x} (e^{x + y} (x - 1) (y - 2))

e^{x + y} x (y - 2)

Solution steps

\frac{\partial}{\partial x} (e^{x + y} (x - 1) (y - 2))

Treat

y

as a constant

Take the constant out:

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

= (y - 2) \frac{\partial}{\partial x} (e^{x + y} (x - 1))

Apply the Product Rule:

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

f = e^{x + y}, g = x - 1

= (y - 2) (\frac{\partial}{\partial x} (e^{x + y}) (x - 1) + \frac{\partial}{\partial x} (x - 1) e^{x + y})

\frac{\partial}{\partial x} (e^{x + y}) = e^{x + y}

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

= (y - 2) (e^{x + y} (x - 1) + 1 \cdot e^{x + y})

Simplify

(y - 2) (e^{x + y} (x - 1) + 1 \cdot e^{x + y}) : e^{x + y} x (y - 2)

= e^{x + y} x (y - 2)

\frac{s + 1}{( s + 1 ) ^{2} + 4}

\int x^{2} 4 x + 8 d x

y = x^{2}, y = 4 + x

\int x (x - 3)^{2} + 4^{2} d x

y = x^{2}, y = x + 2, x = - 2

What is (\partial)/(\partial x)(e^{x+y}(x-1)(y-2)) ?
The answer to (\partial)/(\partial x)(e^{x+y}(x-1)(y-2)) is e^{x+y}x(y-2)