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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\int \frac{6}{( x - 4 ) ( x + 2 )} d x

y^{^{''}} + 81 y = sec^{2} (9 x)

(\frac{1}{4}, \frac{7}{8}), (4 \frac{1}{4}, 2 \frac{7}{8})

n = 0 \sum \infty \frac{1}{( 2 n + 1 ) ^{6}}

y d x + (1 + x) d y = 0

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