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

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

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

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

e^{2 x} \cdot 2 + \frac{y}{x}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= e^{2 x} \cdot 2 + \frac{y}{x}

2 y^{^{''}} + y^{^{'}} - 10 y = 0, y (1) = 5, y^{^{'}} (1) = 2

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 5 x^{3}))

3 y^{^{'}} + 15 y = 180

\int_{0}^{2} \frac{1}{8 + 2 t ^{2}} d t

\int (arccos (x)) d x

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