What is (\partial)/(\partial x)(e^ye^x+ln(xy)) ?

The answer to (\partial)/(\partial x)(e^ye^x+ln(xy)) is e^{y+x}+1/x

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(\partial)/(\partial x)(e^ye^x+ln(xy))

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

e^{y + x} + \frac{1}{x}

Solution steps

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

Simplify

e^{y} e^{x} + ln (x y) : e^{y + x} + ln (y x)

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (ln (y x)) = \frac{1}{x}

= e^{y + x} + \frac{1}{x}

e^{s^{2} + 1}

\frac{d y}{d x} - \frac{1}{x} y = 1 - \frac{1}{x}

f (x) = cos (6 x)

\frac{tan ( 3 x )}{x ^{2}}

\frac{d}{d θ} (sin (θ) + 2 cos (θ))

What is (\partial)/(\partial x)(e^ye^x+ln(xy)) ?
The answer to (\partial)/(\partial x)(e^ye^x+ln(xy)) is e^{y+x}+1/x