Solve (5v^2+6v+4)+(v^2+2v+3)

$$( 5 v ^ { 2 } + 6 v + 4 ) + ( v ^ { 2 } + 2 v + 3 )$$

Evaluate

$$6v^{2}+8v+7$$

Solution Steps

Combine 5v^{2} and v^{2} to get 6v^{2}.

$$6v^{2}+6v+4+2v+3$$

Combine 6v and 2v to get 8v.

$$6v^{2}+8v+4+3$$

Add 4 and 3 to get 7.

$$6v^{2}+8v+7$$

Differentiate w.r.t. v

$$12v+8$$

Steps Using Derivative Rule for Sum

Combine 5v^{2} and v^{2} to get 6v^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}v}(6v^{2}+6v+4+2v+3)$$

Combine 6v and 2v to get 8v.

$$\frac{\mathrm{d}}{\mathrm{d}v}(6v^{2}+8v+4+3)$$

Add 4 and 3 to get 7.

$$\frac{\mathrm{d}}{\mathrm{d}v}(6v^{2}+8v+7)$$

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}.

$$2\times 6v^{2-1}+8v^{1-1}$$

Multiply 2 times 6.

$$12v^{2-1}+8v^{1-1}$$

Subtract 1 from 2.

$$12v^{1}+8v^{1-1}$$

Subtract 1 from 1.

$$12v^{1}+8v^{0}$$

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

$$12v+8v^{0}$$

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

$$12v+8\times 1$$

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

$$12v+8$$