Solve (7k^2+6k+5)+(6k^2+5k+8)

$$( 7 k ^ { 2 } + 6 k + 5 ) + ( 6 k ^ { 2 } + 5 k + 8 )$$

Evaluate

$$13k^{2}+11k+13$$

Solution Steps

Combine 7k^{2} and 6k^{2} to get 13k^{2}.

$$13k^{2}+6k+5+5k+8$$

Combine 6k and 5k to get 11k.

$$13k^{2}+11k+5+8$$

Add 5 and 8 to get 13.

$$13k^{2}+11k+13$$

Differentiate w.r.t. k

$$26k+11$$

Steps Using Derivative Rule for Sum

Combine 7k^{2} and 6k^{2} to get 13k^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(13k^{2}+6k+5+5k+8)$$

Combine 6k and 5k to get 11k.

$$\frac{\mathrm{d}}{\mathrm{d}k}(13k^{2}+11k+5+8)$$

Add 5 and 8 to get 13.

$$\frac{\mathrm{d}}{\mathrm{d}k}(13k^{2}+11k+13)$$

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}.

$$2\times 13k^{2-1}+11k^{1-1}$$

Multiply 2 times 13.

$$26k^{2-1}+11k^{1-1}$$

Subtract 1 from 2.

$$26k^{1}+11k^{1-1}$$

Subtract 1 from 1.

$$26k^{1}+11k^{0}$$

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

$$26k+11k^{0}$$

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

$$26k+11\times 1$$

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

$$26k+11$$