Solve 3/5b-2+7b-12-2/5b+14

$$\frac { 3 } { 5 } b - 2 + 7 b - 12 - \frac { 2 } { 5 } b + 14$$

Evaluate

$$\frac{36b}{5}$$

Solution Steps

Combine \frac{3}{5}b and 7b to get \frac{38}{5}b.

$$\frac{38}{5}b-2-12-\frac{2}{5}b+14$$

Subtract 12 from -2 to get -14.

$$\frac{38}{5}b-14-\frac{2}{5}b+14$$

Combine \frac{38}{5}b and -\frac{2}{5}b to get \frac{36}{5}b.

$$\frac{36}{5}b-14+14$$

Add -14 and 14 to get 0.

$$\frac{36}{5}b$$

Differentiate w.r.t. b

$$\frac{36}{5} = 7\frac{1}{5} = 7.2$$

Steps Using Definition of a Derivative

Combine \frac{3}{5}b and 7b to get \frac{38}{5}b.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{38}{5}b-2-12-\frac{2}{5}b+14)$$

Subtract 12 from -2 to get -14.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{38}{5}b-14-\frac{2}{5}b+14)$$

Combine \frac{38}{5}b and -\frac{2}{5}b to get \frac{36}{5}b.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{36}{5}b-14+14)$$

Add -14 and 14 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{36}{5}b)$$

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

$$\frac{36}{5}b^{1-1}$$

Subtract 1 from 1.

$$\frac{36}{5}b^{0}$$

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

$$\frac{36}{5}\times 1$$

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

$$\frac{36}{5}$$