What is (\partial)/(\partial x)(e^xln(xy)) ?

The answer to (\partial)/(\partial x)(e^xln(xy)) is e^xln(xy)+(e^x)/x

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(\partial)/(\partial x)(e^xln(xy))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{x}, g = ln (x y)

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

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

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

= e^{x} ln (x y) + \frac{1}{x} e^{x}

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

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

2 x^{3} - 3 x^{2} - 12 + 8

n = 1 \sum \infty 7^{n} x^{n - 1}

y^{^{''}} + 6 y^{^{'}} + 8 y = cos (3 t), y (0) = - 2, y^{^{'}} (0) = 1

\int_{6}^{8} \frac{86}{( x - 6 ) ^{3}} d x

\frac{\partial}{\partial x} (2 x^{4} y^{2} + 3 x^{2} y^{4})

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