Solve (8p^2+3p+9)+(4p^2+p+2)

$$( 8 p ^ { 2 } + 3 p + 9 ) + ( 4 p ^ { 2 } + p + 2 )$$

Evaluate

$$12p^{2}+4p+11$$

Solution Steps

Combine 8p^{2} and 4p^{2} to get 12p^{2}.

$$12p^{2}+3p+9+p+2$$

Combine 3p and p to get 4p.

$$12p^{2}+4p+9+2$$

Add 9 and 2 to get 11.

$$12p^{2}+4p+11$$

Differentiate w.r.t. p

$$24p+4$$

Steps Using Derivative Rule for Sum

Combine 8p^{2} and 4p^{2} to get 12p^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}p}(12p^{2}+3p+9+p+2)$$

Combine 3p and p to get 4p.

$$\frac{\mathrm{d}}{\mathrm{d}p}(12p^{2}+4p+9+2)$$

Add 9 and 2 to get 11.

$$\frac{\mathrm{d}}{\mathrm{d}p}(12p^{2}+4p+11)$$

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}.

$$2\times 12p^{2-1}+4p^{1-1}$$

Multiply 2 times 12.

$$24p^{2-1}+4p^{1-1}$$

Subtract 1 from 2.

$$24p^{1}+4p^{1-1}$$

Subtract 1 from 1.

$$24p^{1}+4p^{0}$$

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

$$24p+4p^{0}$$

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

$$24p+4\times 1$$

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

$$24p+4$$