Solve 5b^2+9b+10+3b+2b^2-4

$$5 b ^ { 2 } + 9 b + 10 + 3 b + 2 b ^ { 2 } - 4$$

Evaluate

$$7b^{2}+12b+6$$

Solution Steps

Combine 9b and 3b to get 12b.

$$5b^{2}+12b+10+2b^{2}-4$$

Combine 5b^{2} and 2b^{2} to get 7b^{2}.

$$7b^{2}+12b+10-4$$

Subtract 4 from 10 to get 6.

$$7b^{2}+12b+6$$

Differentiate w.r.t. b

$$14b+12$$

Steps Using Derivative Rule for Sum

Combine 9b and 3b to get 12b.

$$\frac{\mathrm{d}}{\mathrm{d}b}(5b^{2}+12b+10+2b^{2}-4)$$

Combine 5b^{2} and 2b^{2} to get 7b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}b}(7b^{2}+12b+10-4)$$

Subtract 4 from 10 to get 6.

$$\frac{\mathrm{d}}{\mathrm{d}b}(7b^{2}+12b+6)$$

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}.

$$2\times 7b^{2-1}+12b^{1-1}$$

Multiply 2 times 7.

$$14b^{2-1}+12b^{1-1}$$

Subtract 1 from 2.

$$14b^{1}+12b^{1-1}$$

Subtract 1 from 1.

$$14b^{1}+12b^{0}$$

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

$$14b+12b^{0}$$

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

$$14b+12\times 1$$

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

$$14b+12$$