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

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

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

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

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

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

\frac{\partial}{\partial y} (3 x^{2} y^{4} - 4)

\frac{2}{x ^{2} + 3}

2 a x + b

\frac{d y}{d x} = \frac{2 x}{y}

\int_{- 8}^{8} x^{\frac{1}{3}} d x

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