Solve (2v^2+v+6)+(4v^2+3v+7)

$$( 2 v ^ { 2 } + v + 6 ) + ( 4 v ^ { 2 } + 3 v + 7 )$$

Evaluate

$$6v^{2}+4v+13$$

Solution Steps

Combine 2v^{2} and 4v^{2} to get 6v^{2}.

$$6v^{2}+v+6+3v+7$$

Combine v and 3v to get 4v.

$$6v^{2}+4v+6+7$$

Add 6 and 7 to get 13.

$$6v^{2}+4v+13$$

Differentiate w.r.t. v

$$12v+4$$

Steps Using Derivative Rule for Sum

Combine 2v^{2} and 4v^{2} to get 6v^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}v}(6v^{2}+v+6+3v+7)$$

Combine v and 3v to get 4v.

$$\frac{\mathrm{d}}{\mathrm{d}v}(6v^{2}+4v+6+7)$$

Add 6 and 7 to get 13.

$$\frac{\mathrm{d}}{\mathrm{d}v}(6v^{2}+4v+13)$$

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}.

$$2\times 6v^{2-1}+4v^{1-1}$$

Multiply 2 times 6.

$$12v^{2-1}+4v^{1-1}$$

Subtract 1 from 2.

$$12v^{1}+4v^{1-1}$$

Subtract 1 from 1.

$$12v^{1}+4v^{0}$$

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

$$12v+4v^{0}$$

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

$$12v+4\times 1$$

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

$$12v+4$$