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

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

Popular >

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

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

y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

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

= 0 + y + 0 - 0

Simplify

= y

\int \frac{5}{x ^{2} - 1} d x

\frac{d}{d x} (2^{x} + x)

x \to 2 lim (\frac{1}{x ^{2}})

x \to π lim (\frac{x}{3} - \frac{2 π}{3})

\frac{d ^{2} x}{d t ^{2}} + 4 x = t^{2}

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