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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

\frac{\partial}{\partial y} (y) = 1

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

= - 2 x (1 \cdot sin (x^{2} y) + cos (x^{2} y) x^{2} y)

Simplify

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

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