Solve (k^3+6k^2+8k+5)+(3k+6)

$$( k ^ { 3 } + 6 k ^ { 2 } + 8 k + 5 ) + ( 3 k + 6 )$$

Evaluate

$$k^{3}+6k^{2}+11k+11$$

Solution Steps

Combine 8k and 3k to get 11k.

$$k^{3}+6k^{2}+11k+5+6$$

Add 5 and 6 to get 11.

$$k^{3}+6k^{2}+11k+11$$

Differentiate w.r.t. k

$$3k^{2}+12k+11$$

Steps Using Derivative Rule for Sum

Combine 8k and 3k to get 11k.

$$\frac{\mathrm{d}}{\mathrm{d}k}(k^{3}+6k^{2}+11k+5+6)$$

Add 5 and 6 to get 11.

$$\frac{\mathrm{d}}{\mathrm{d}k}(k^{3}+6k^{2}+11k+11)$$

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}.

$$3k^{3-1}+2\times 6k^{2-1}+11k^{1-1}$$

Subtract 1 from 3.

$$3k^{2}+2\times 6k^{2-1}+11k^{1-1}$$

Multiply 2 times 6.

$$3k^{2}+12k^{2-1}+11k^{1-1}$$

Subtract 1 from 2.

$$3k^{2}+12k^{1}+11k^{1-1}$$

Subtract 1 from 1.

$$3k^{2}+12k^{1}+11k^{0}$$

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

$$3k^{2}+12k+11k^{0}$$

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

$$3k^{2}+12k+11\times 1$$

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

$$3k^{2}+12k+11$$