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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

y^{^{''}} + 8 y^{^{'}} + \frac{73}{4} y = 0

\int 2 \cdot e^{- 2 x} d x

x \to 6 - lim (\frac{∣ x - 6 ∣}{x - 6})

f (θ) = cot^{2} (sin (θ))

\frac{\partial}{\partial y} (ln (x^{2} + y^{4}))

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