What is (\partial)/(\partial x)(x*e^y+ln(x)*y) ?

The answer to (\partial)/(\partial x)(x*e^y+ln(x)*y) is e^y+y/x

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(\partial)/(\partial x)(xe^y+ln(x)y)

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{v}{v + \frac{c}{v}}

\int_{0}^{1} e^{- 5 x^{2}} d x

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

n = 1 \sum \infty 1 (\frac{1}{6})^{n}

n = 1 \sum \infty n^{n} (x - 3)^{n}

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