Solve (d^2+6d+9)+(d^3+6d+9)=

$$( d ^ { 2 } + 6 d + 9 ) + ( d ^ { 3 } + 6 d + 9 ) =$$

Evaluate

$$d^{3}+d^{2}+12d+18$$

Solution Steps

Combine 6d and 6d to get 12d.

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

Add 9 and 9 to get 18.

$$d^{2}+12d+18+d^{3}$$

Differentiate w.r.t. d

$$3d^{2}+2d+12$$

Steps Using Derivative Rule for Sum

Combine 6d and 6d to get 12d.

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

Add 9 and 9 to get 18.

$$\frac{\mathrm{d}}{\mathrm{d}d}(d^{2}+12d+18+d^{3})$$

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}.

$$2d^{2-1}+12d^{1-1}+3d^{3-1}$$

Subtract 1 from 2.

$$2d^{1}+12d^{1-1}+3d^{3-1}$$

Subtract 1 from 1.

$$2d^{1}+12d^{0}+3d^{3-1}$$

Subtract 1 from 3.

$$2d^{1}+12d^{0}+3d^{2}$$

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

$$2d+12d^{0}+3d^{2}$$

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

$$2d+12\times 1+3d^{2}$$

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

$$2d+12+3d^{2}$$