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

\frac{d}{d x} (\frac{3 x + 1}{x + 1})

\frac{2}{s ^{2} - 2 s + 9}

y^{^{''}} + 2 y^{^{'}} + y = x^{- 4} e^{- x}

y = \frac{x}{x ^{2} + 9}

x \to \infty lim (1 + \frac{2}{x})

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