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

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

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

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

- x^{2} cos (x y)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= - x cos (x y) x

Simplify

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

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

\frac{\partial}{\partial y} (5 x (4 + y)^{- 1})

5 y^{^{''}} - 6 y^{^{'}} + 5 y = 0

f (x) = \frac{16}{x}

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

f (x) = sin^{2} (x^{2})

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