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

The answer to (\partial)/(\partial y)(cos(x^2+y^2)) is -sin(x^2+y^2)*2y

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

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

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

Solution steps

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

Treat

x

as a constant

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

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

\int (3 x + 5) cos (\frac{x}{4}) d x

\frac{\partial}{\partial y} (3 x z + y^{2})

e^{\frac{x}{y}}

3 (e^{x} cos (x) - e^{x} sin (x))

x \to 0 lim (\frac{6 x + 4 - 2}{x})

What is (\partial)/(\partial y)(cos(x^2+y^2)) ?
The answer to (\partial)/(\partial y)(cos(x^2+y^2)) is -sin(x^2+y^2)*2y