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

The answer to (\partial)/(\partial y)(sin(x^2-y^2)) is -2ycos(x^2-y^2)

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

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

- 2 y cos (x^{2} - y^{2})

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}) (- 2 y)

Simplify

= - 2 y cos (x^{2} - y^{2})

cos (e^{x})

\int_{1}^{\infty} 3^{- x} d x

\int_{0}^{3} 5 d x

\int - ln (1 - x) d x

x + y = 2

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