Solve (12+8k^3+3k-4k^2)+(5k^3+15-k+2k^2)

$$( 12 + 8 k ^ { 3 } + 3 k - 4 k ^ { 2 } ) + ( 5 k ^ { 3 } + 15 - k + 2 k ^ { 2 } )$$

Evaluate

$$13k^{3}-2k^{2}+2k+27$$

Solution Steps

Combine 8k^{3} and 5k^{3} to get 13k^{3}.

$$12+13k^{3}+3k-4k^{2}+15-k+2k^{2}$$

Add 12 and 15 to get 27.

$$27+13k^{3}+3k-4k^{2}-k+2k^{2}$$

Combine 3k and -k to get 2k.

$$27+13k^{3}+2k-4k^{2}+2k^{2}$$

Combine -4k^{2} and 2k^{2} to get -2k^{2}.

$$27+13k^{3}+2k-2k^{2}$$

Differentiate w.r.t. k

$$39k^{2}-4k+2$$

Steps Using Derivative Rule for Sum

Combine 8k^{3} and 5k^{3} to get 13k^{3}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(12+13k^{3}+3k-4k^{2}+15-k+2k^{2})$$

Add 12 and 15 to get 27.

$$\frac{\mathrm{d}}{\mathrm{d}k}(27+13k^{3}+3k-4k^{2}-k+2k^{2})$$

Combine 3k and -k to get 2k.

$$\frac{\mathrm{d}}{\mathrm{d}k}(27+13k^{3}+2k-4k^{2}+2k^{2})$$

Combine -4k^{2} and 2k^{2} to get -2k^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}k}(27+13k^{3}+2k-2k^{2})$$

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}.

$$3\times 13k^{3-1}+2k^{1-1}+2\left(-2\right)k^{2-1}$$

Multiply 3 times 13.

$$39k^{3-1}+2k^{1-1}+2\left(-2\right)k^{2-1}$$

Subtract 1 from 3.

$$39k^{2}+2k^{1-1}+2\left(-2\right)k^{2-1}$$

Subtract 1 from 1.

$$39k^{2}+2k^{0}+2\left(-2\right)k^{2-1}$$

Multiply 1 times 2.

$$39k^{2}+2k^{0}-4k^{2-1}$$

Subtract 1 from 2.

$$39k^{2}+2k^{0}-4k^{1}$$

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

$$39k^{2}+2k^{0}-4k$$

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

$$39k^{2}+2\times 1-4k$$

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

$$39k^{2}+2-4k$$