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

\frac{d}{d x} (\frac{x + 3}{x - 4})

y = \frac{9 x ^{2} + 8}{x ^{2} + 1}

\int (\frac{6 x ^{5} - 3 x ^{2} + 4}{x ^{2}}) d x

\frac{d}{d x} (e^{s i n (x)})

- \frac{2 d y}{d x} + 6 x^{2} = 0

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