Solve b|(-1)(-6)(+2)-3+8=

$$b | ( - 1 ) ( - 6 ) ( + 2 ) - 3 + 8 =$$

Evaluate

$$17b$$

Solution Steps

Multiply -1 and -6 to get 6.

$$b|6\times 2-3+8|$$

Multiply 6 and 2 to get 12.

$$b|12-3+8|$$

Subtract 3 from 12 to get 9.

$$b|9+8|$$

Add 9 and 8 to get 17.

$$b|17|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 17 is 17.

$$b\times 17$$

Differentiate w.r.t. b

$$17$$

Steps Using Definition of a Derivative

Multiply -1 and -6 to get 6.

$$\frac{\mathrm{d}}{\mathrm{d}b}(b|6\times 2-3+8|)$$

Multiply 6 and 2 to get 12.

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

Subtract 3 from 12 to get 9.

$$\frac{\mathrm{d}}{\mathrm{d}b}(b|9+8|)$$

Add 9 and 8 to get 17.

$$\frac{\mathrm{d}}{\mathrm{d}b}(b|17|)$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 17 is 17.

$$\frac{\mathrm{d}}{\mathrm{d}b}(b\times 17)$$

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

$$17b^{1-1}$$

Subtract 1 from 1.

$$17b^{0}$$

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

$$17\times 1$$

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

$$17$$