Solve -21x^2y^4div(-3x^2y^3)

$$- 21 x ^ { 2 } y ^ { 4 } \div ( - 3 x ^ { 2 } y ^ { 3 } )$$

Evaluate

$$7y$$

Steps Using Quotient of Powers Property

Use the rules of exponents to simplify the expression.

$$\frac{\left(-21\right)^{1}x^{2}y^{4}}{\left(-3\right)^{1}x^{2}y^{3}}$$

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent.

$$\frac{\left(-21\right)^{1}}{\left(-3\right)^{1}}x^{2-2}y^{4-3}$$

Subtract 2 from 2.

$$\frac{\left(-21\right)^{1}}{\left(-3\right)^{1}}x^{0}y^{4-3}$$

For any number a except 0, a^{0}=1.

$$\frac{\left(-21\right)^{1}}{\left(-3\right)^{1}}y^{4-3}$$

Subtract 3 from 4.

$$\frac{\left(-21\right)^{1}}{\left(-3\right)^{1}}y^{1}$$

Divide -21 by -3.

$$7y$$

Differentiate w.r.t. y

$$7$$

Steps Using Definition of a Derivative

Cancel out 3x^{2}y^{3} in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}y}(\frac{-7y}{-1})$$

Anything divided by -1 gives its opposite.

$$\frac{\mathrm{d}}{\mathrm{d}y}(7y)$$

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

$$7y^{1-1}$$

Subtract 1 from 1.

$$7y^{0}$$

For any term t except 0, t^{0}=1.

$$7\times 1$$

For any term t, t\times 1=t and 1t=t.

$$7$$