Solve (b+9)+(b+3)+(b+7

$$( b + 9 ) + ( b + 3 ) + ( b + 7$$

Evaluate

$$3b+19$$

Solution Steps

Combine b and b to get 2b.

$$2b+9+3+b+7$$

Add 9 and 3 to get 12.

$$2b+12+b+7$$

Combine 2b and b to get 3b.

$$3b+12+7$$

Add 12 and 7 to get 19.

$$3b+19$$

Differentiate w.r.t. b

$$3$$

Steps Using Derivative Rule for Sum

Combine b and b to get 2b.

$$\frac{\mathrm{d}}{\mathrm{d}b}(2b+9+3+b+7)$$

Add 9 and 3 to get 12.

$$\frac{\mathrm{d}}{\mathrm{d}b}(2b+12+b+7)$$

Combine 2b and b to get 3b.

$$\frac{\mathrm{d}}{\mathrm{d}b}(3b+12+7)$$

Add 12 and 7 to get 19.

$$\frac{\mathrm{d}}{\mathrm{d}b}(3b+19)$$

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}.

$$3b^{1-1}$$

Subtract 1 from 1.

$$3b^{0}$$

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

$$3\times 1$$

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

$$3$$