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

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

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(\partial)/(\partial x)(-sin(xy)y)

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

- y^{2} cos (x y)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

cos (x y) \frac{\partial}{\partial x} (x y)

= cos (x y) \frac{\partial}{\partial x} (x y)

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

= - y cos (x y) y

Simplify

- y cos (x y) y : - y^{2} cos (x y)

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

y^{^{'}} - x y^{2} = 2 x y

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

\int x^{2} sin (9 x) d x

\int \frac{x ^{2} + 1}{3 x ^{2} - x - 2} d x

t \to 0 + lim (t (ln (t))^{2})

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