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

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

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

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

x^{2} (3 y^{2} sin (x - 2 y) - 2 y^{3} cos (x - 2 y))

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y^{3}, g = sin (x - 2 y)

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

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

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

= x^{2} (3 y^{2} sin (x - 2 y) + (- 2 cos (x - 2 y)) y^{3})

Simplify

= x^{2} (3 y^{2} sin (x - 2 y) - 2 y^{3} cos (x - 2 y))

x \to 7 lim (x + 6)

y^{^{'}} = \frac{x}{14 y}

\int_{\frac{3 π}{2}}^{2 π} - 4 sin (x) d x

x \to - 2 lim (ln (x + 2))

\frac{d}{d x} (\frac{x}{x - 4})

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