Solve -8p^2q^3/-32p^3q^3=

$$\frac { - 8 p ^ { 2 } q ^ { 3 } } { - 32 p ^ { 3 } q ^ { 3 } } =$$

Evaluate

$$\frac{1}{4p}$$

Steps Using Quotient of Powers Property

Use the rules of exponents to simplify the expression.

$$\frac{\left(-8\right)^{1}p^{2}q^{3}}{\left(-32\right)^{1}p^{3}q^{3}}$$

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

$$\frac{\left(-8\right)^{1}}{\left(-32\right)^{1}}p^{2-3}q^{3-3}$$

Subtract 3 from 2.

$$\frac{\left(-8\right)^{1}}{\left(-32\right)^{1}}\times \frac{1}{p}q^{3-3}$$

Subtract 3 from 3.

$$\frac{\left(-8\right)^{1}}{\left(-32\right)^{1}}\times \frac{1}{p}q^{0}$$

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

$$\frac{\left(-8\right)^{1}}{\left(-32\right)^{1}}\times \frac{1}{p}$$

Reduce the fraction \frac{-8}{-32} to lowest terms by extracting and canceling out 8.

$$\frac{1}{4}\times \frac{1}{p}$$

Differentiate w.r.t. p

$$-\frac{1}{4p^{2}}$$

Solution Steps

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

$$\frac{\mathrm{d}}{\mathrm{d}p}(\left(-\frac{8q^{3}}{-32q^{3}}\right)p^{2-3})$$

Do the arithmetic.

$$\frac{\mathrm{d}}{\mathrm{d}p}(\frac{1}{4}\times \frac{1}{p})$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$-\frac{1}{4}p^{-1-1}$$

Do the arithmetic.

$$-\frac{1}{4}p^{-2}$$