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

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

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

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

cos (x^{2} + y^{2}) \cdot 2 y

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

\frac{d y}{d x} = \frac{2 x ( y - 1 )}{x ^{2} + 3}

\int \frac{1 + cos ( x )}{1 + sin ( x )} d x

\frac{\partial}{\partial x} (5 x^{4} + 2 x y + 1)

e^{2 t} cos (3 t)

\int + \frac{x ^{4} + 12 x ^{2} + 1}{x ^{5} + 20 x ^{3} + 5 x} d x

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