What is (\partial)/(\partial y)((2x+y-4)e^{2y}) ?

The answer to (\partial)/(\partial y)((2x+y-4)e^{2y}) is -7e^{2y}+4e^{2y}x+2e^{2y}y

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(\partial)/(\partial y)((2x+y-4)e^{2y})

\frac{\partial}{\partial y} ((2 x + y - 4) e^{2 y})

- 7 e^{2 y} + 4 e^{2 y} x + 2 e^{2 y} y

Solution steps

\frac{\partial}{\partial y} ((2 x + y - 4) e^{2 y})

Treat

x

as a constant

Apply the Product Rule:

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

f = 2 x + y - 4, g = e^{2 y}

= \frac{\partial}{\partial y} (2 x + y - 4) e^{2 y} + \frac{\partial}{\partial y} (e^{2 y}) (2 x + y - 4)

\frac{\partial}{\partial y} (2 x + y - 4) = 1

\frac{\partial}{\partial y} (e^{2 y}) = e^{2 y} \cdot 2

= 1 \cdot e^{2 y} + e^{2 y} \cdot 2 (2 x + y - 4)

Simplify

1 \cdot e^{2 y} + e^{2 y} \cdot 2 (2 x + y - 4) : - 7 e^{2 y} + 4 e^{2 y} x + 2 e^{2 y} y

= - 7 e^{2 y} + 4 e^{2 y} x + 2 e^{2 y} y

\int sin (t) cos (t) d t

x \to 0.1 lim (\frac{8 sin ( x )}{x})

2 x^{4} 25 - x^{2}

y^{^{''''}} + 1 = 0

\frac{\partial}{\partial x} (\frac{4}{x y} - x^{2} + 3 y^{2})

What is (\partial)/(\partial y)((2x+y-4)e^{2y}) ?
The answer to (\partial)/(\partial y)((2x+y-4)e^{2y}) is -7e^{2y}+4e^{2y}x+2e^{2y}y