What is (\partial)/(\partial y)(sin(x^3+y^2)) ?

The answer to (\partial)/(\partial y)(sin(x^3+y^2)) is cos(x^3+y^2)*2y

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(\partial)/(\partial y)(sin(x^3+y^2))

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

cos (x^{3} + y^{2}) \cdot 2 y

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

= cos (x^{3} + y^{2}) \cdot 2 y

What is (\partial)/(\partial y)(sin(x^3+y^2)) ?
The answer to (\partial)/(\partial y)(sin(x^3+y^2)) is cos(x^3+y^2)*2y