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

The answer to (\partial)/(\partial y)(2xcos(x^2+y^2)) is -4xysin(x^2+y^2)

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

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

- 4 x y sin (x^{2} + y^{2})

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

= 2 x (- sin (x^{2} + y^{2}) \cdot 2 y)

Simplify

= - 4 x y sin (x^{2} + y^{2})

y = e^{- 2 x^{3}}

\frac{x ^{2}}{3 + x}

\frac{d y}{d x} + y cos (x) = 5 cos (x), y (0) = 7

\frac{d}{d x} (x^{7})

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

What is (\partial)/(\partial y)(2xcos(x^2+y^2)) ?
The answer to (\partial)/(\partial y)(2xcos(x^2+y^2)) is -4xysin(x^2+y^2)