What is (\partial)/(\partial y)(e^{2x}+yln(x)) ?

The answer to (\partial)/(\partial y)(e^{2x}+yln(x)) is ln(x)

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(\partial)/(\partial y)(e^{2x}+yln(x))

\frac{\partial}{\partial y} (e^{2 x} + y ln (x))

ln (x)

Solution steps

\frac{\partial}{\partial y} (e^{2 x} + y ln (x))

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (e^{2 x}) + \frac{\partial}{\partial y} (y ln (x))

\frac{\partial}{\partial y} (e^{2 x}) = 0

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

= 0 + ln (x)

Simplify

= ln (x)

\frac{\partial}{\partial x} (x arctan (\frac{y}{z}))

\frac{d}{d x} (\frac{e ^{3 x}}{x ^{15}})

y = 2 x + 1, y = 13 - x^{2}, x = - 1, x = 2

\int \frac{ln ( 9 x )}{x} d x

\int (1 - cos^{2} (x))^{2} sin (x) d x

What is (\partial)/(\partial y)(e^{2x}+yln(x)) ?
The answer to (\partial)/(\partial y)(e^{2x}+yln(x)) is ln(x)