What is the d/(dy)((x(y))(y^2-1)) ?

The d/(dy)((x(y))(y^2-1)) is x(3y^2-1)

What is the first d/(dy)((x(y))(y^2-1)) ?

The first d/(dy)((x(y))(y^2-1)) is x(3y^2-1)

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\frac{d}{d y} ((x (y)) (y^{2} - 1))

x (3 y^{2} - 1)

Solution steps

\frac{d}{d y} ((x (y)) (y^{2} - 1))

Treat

x

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = y, g = y^{2} - 1

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

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

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

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

Simplify

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

= x (3 y^{2} - 1)

(3 i, - 1 j, - 1 k) \cdot (2 i, j, 2 k)

[(x^{2} + 2) (x - 1)]^{^{'}}

\frac{3 ( x + h ) + 7}{7 ( x + h ) - 8}

\frac{3 ( 7 - 2 3 )}{37}

(x + 1) ln (x) (1, e)

What is the d/(dy)((x(y))(y^2-1)) ?
The d/(dy)((x(y))(y^2-1)) is x(3y^2-1)
What is the first d/(dy)((x(y))(y^2-1)) ?
The first d/(dy)((x(y))(y^2-1)) is x(3y^2-1)