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

The d/(dy)((x(y))(y^2+y)) is x(3y^2+2y)

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

The first d/(dy)((x(y))(y^2+y)) is x(3y^2+2y)

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

x (3 y^{2} + 2 y)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = y, g = y^{2} + y

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

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

\frac{d}{d y} (y^{2} + y) = 2 y + 1

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

Simplify

x (1 \cdot (y^{2} + y) + (2 y + 1) y) : x (3 y^{2} + 2 y)

= x (3 y^{2} + 2 y)

[2 + 2] (2 + 1)^{k}

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

- \frac{6 ( s + 1 )}{( s ^{2} + 5 s + 6 )}

(2, - 2) y (5, - 3)

(- 2 x)^{\circ - 1} (x + y)

What is the d/(dy)((x(y))(y^2+y)) ?
The d/(dy)((x(y))(y^2+y)) is x(3y^2+2y)
What is the first d/(dy)((x(y))(y^2+y)) ?
The first d/(dy)((x(y))(y^2+y)) is x(3y^2+2y)