Solve j|-2/3-1/4|

$$j | - \frac { 2 } { 3 } - \frac { 1 } { 4 } |$$

Evaluate

$$\frac{11j}{12}$$

Solution Steps

Least common multiple of 3 and 4 is 12. Convert -\frac{2}{3} and \frac{1}{4} to fractions with denominator 12.

$$j|-\frac{8}{12}-\frac{3}{12}|$$

Since -\frac{8}{12} and \frac{3}{12} have the same denominator, subtract them by subtracting their numerators.

$$j|\frac{-8-3}{12}|$$

Subtract 3 from -8 to get -11.

$$j|-\frac{11}{12}|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{11}{12} is \frac{11}{12}.

$$j\times \frac{11}{12}$$

Differentiate w.r.t. j

$$\frac{11}{12} = 0.9166666666666666$$

Steps Using Definition of a Derivative

Least common multiple of 3 and 4 is 12. Convert -\frac{2}{3} and \frac{1}{4} to fractions with denominator 12.

$$\frac{\mathrm{d}}{\mathrm{d}j}(j|-\frac{8}{12}-\frac{3}{12}|)$$

Since -\frac{8}{12} and \frac{3}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}j}(j|\frac{-8-3}{12}|)$$

Subtract 3 from -8 to get -11.

$$\frac{\mathrm{d}}{\mathrm{d}j}(j|-\frac{11}{12}|)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -\frac{11}{12} is \frac{11}{12}.

$$\frac{\mathrm{d}}{\mathrm{d}j}(j\times \frac{11}{12})$$

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

$$\frac{11}{12}j^{1-1}$$

Subtract 1 from 1.

$$\frac{11}{12}j^{0}$$

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

$$\frac{11}{12}\times 1$$

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

$$\frac{11}{12}$$