How to solve 2y * 3y * 4y =

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 2y * 3y * 4y = | Equatix

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Question

$$2y\times3y\times4y=$$

Evaluate

$24y^{3}$

Solution Steps

Multiply $y$ and $y$ to get $y^{2}$.

$$2y^{2}\times 3\times 4y$$

To multiply powers of the same base, add their exponents. Add $2$ and $1$ to get $3$.

$$2y^{3}\times 3\times 4$$

Multiply $2$ and $3$ to get $6$.

$$6y^{3}\times 4$$

Multiply $6$ and $4$ to get $24$.

$$24y^{3}$$

Show Solution

Differentiate w.r.t. y

$72y^{2}$

Steps Using Definition of a Derivative

Multiply $y$ and $y$ to get $y^{2}$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(2y^{2}\times 3\times 4y)$$

To multiply powers of the same base, add their exponents. Add $2$ and $1$ to get $3$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(2y^{3}\times 3\times 4)$$

Multiply $2$ and $3$ to get $6$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(6y^{3}\times 4)$$

Multiply $6$ and $4$ to get $24$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(24y^{3})$$

The derivative of $ax^{n}$ is $nax^{n-1}$.

$$3\times 24y^{3-1}$$

Multiply $3$ times $24$.

$$72y^{3-1}$$

Subtract $1$ from $3$.

$$72y^{2}$$