What is the d/(dy)(-2xysin(x^2y)) ?

The d/(dy)(-2xysin(x^2y)) is -2x(sin(x^2y)+x^2ycos(x^2y))

What is the first d/(dy)(-2xysin(x^2y)) ?

The first d/(dy)(-2xysin(x^2y)) is -2x(sin(x^2y)+x^2ycos(x^2y))

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

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

Solution steps

\frac{d}{d 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{d}{d 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{d y}{d y} sin (x^{2} y) + \frac{d}{d y} (sin (x^{2} y)) y)

\frac{d y}{d y} = 1

\frac{d}{d 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))

\int \frac{1}{sin ^{2} ( θ )} d θ

\int \frac{1}{12 x + 0.02 x ^{2}} d x

\frac{d}{d x} (sin (ln ((x^{2} + 4)^{4})))

7 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

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

What is the d/(dy)(-2xysin(x^2y)) ?
The d/(dy)(-2xysin(x^2y)) is -2x(sin(x^2y)+x^2ycos(x^2y))
What is the first d/(dy)(-2xysin(x^2y)) ?
The first d/(dy)(-2xysin(x^2y)) is -2x(sin(x^2y)+x^2ycos(x^2y))