Solve +7/4a^3b^2:(-21/4a^2b)=

$$+ \frac { 7 } { 4 } a ^ { 3 } b ^ { 2 } : ( - \frac { 21 } { 4 } a ^ { 2 } b ) =$$

Evaluate

$$-\frac{ab}{3}$$

Steps Using Quotient of Powers Property

Use the rules of exponents to simplify the expression.

$$\frac{\left(\frac{7}{4}\right)^{1}a^{3}b^{2}}{\left(-\frac{21}{4}\right)^{1}a^{2}b^{1}}$$

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

$$\frac{\left(\frac{7}{4}\right)^{1}}{\left(-\frac{21}{4}\right)^{1}}a^{3-2}b^{2-1}$$

Subtract 2 from 3.

$$\frac{\left(\frac{7}{4}\right)^{1}}{\left(-\frac{21}{4}\right)^{1}}a^{1}b^{2-1}$$

Subtract 1 from 2.

$$\frac{\left(\frac{7}{4}\right)^{1}}{\left(-\frac{21}{4}\right)^{1}}ab^{1}$$

Divide \frac{7}{4} by -\frac{21}{4} by multiplying \frac{7}{4} by the reciprocal of -\frac{21}{4}.

$$-\frac{1}{3}ab$$

Differentiate w.r.t. a

$$-\frac{b}{3}$$

Solution Steps

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

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{7b^{2}}{4\left(-\frac{21b}{4}\right)}a^{3-2})$$

Do the arithmetic.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\left(-\frac{b}{3}\right)a^{1})$$

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}.

$$\left(-\frac{b}{3}\right)a^{1-1}$$

Do the arithmetic.

$$\left(-\frac{b}{3}\right)a^{0}$$

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

$$\left(-\frac{b}{3}\right)\times 1$$

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

$$-\frac{b}{3}$$