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

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

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(\partial)/(\partial x)(e^y-ye^x)

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

- e^{x} y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - y e^{x}

Simplify

= - e^{x} y

(a - 1) + 1

y^{^{'}} + \frac{2 y}{t + 2} = (t + 2)^{5}, y (0) = 5

\frac{1}{s ^{3} + 4 s}

\int 7 x^{5} e^{x} d x

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