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

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

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

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

e^{x} y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + y e^{x}

Simplify

= e^{x} y

\int \frac{1}{x ^{2} + 6 x + 6} d x

\frac{d y}{d x} + 2 x y = x^{3}

\frac{\partial}{\partial y} (\frac{- y}{x ^{2} + 9 y ^{2}})

\int_{- 2}^{3} \frac{40}{x ^{4}} d x

y = x + 3

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