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

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

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(\partial)/(\partial x)(-sin(x^2y)*2xy)

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\int \frac{x}{1 + 4 x ^{2}} d x

\frac{d}{d x} (\frac{x ^{2}}{2 x + 3})

\int_{0}^{1} \frac{65}{x ^{5}} d x

y^{^{'}} + 4 x y = 0

\int 4 (2 x + 5)^{3} d x

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