What is (\partial)/(\partial y)(x^2+ye^x) ?

The answer to (\partial)/(\partial y)(x^2+ye^x) is e^x

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(\partial)/(\partial y)(x^2+ye^x)

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

e^{x}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + e^{x}

Simplify

= e^{x}

\frac{\partial}{\partial y} (ln (x - 6 y))

(1.85) (5.1)

\int x 4^{x^{2} + 5} d x

\frac{\partial}{\partial x} (4 x^{2} - x)

f (x) = x + 2, at x = 3

What is (\partial)/(\partial y)(x^2+ye^x) ?
The answer to (\partial)/(\partial y)(x^2+ye^x) is e^x