Solve (n+9)+(4n^3+2n^2+4n+2)

$$( n + 9 ) + ( 4 n ^ { 3 } + 2 n ^ { 2 } + 4 n + 2 )$$

Evaluate

$$4n^{3}+2n^{2}+5n+11$$

Solution Steps

Combine n and 4n to get 5n.

$$5n+9+4n^{3}+2n^{2}+2$$

Add 9 and 2 to get 11.

$$5n+11+4n^{3}+2n^{2}$$

Differentiate w.r.t. n

$$12n^{2}+4n+5$$

Steps Using Derivative Rule for Sum

Combine n and 4n to get 5n.

$$\frac{\mathrm{d}}{\mathrm{d}n}(5n+9+4n^{3}+2n^{2}+2)$$

Add 9 and 2 to get 11.

$$\frac{\mathrm{d}}{\mathrm{d}n}(5n+11+4n^{3}+2n^{2})$$

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}.

$$5n^{1-1}+3\times 4n^{3-1}+2\times 2n^{2-1}$$

Subtract 1 from 1.

$$5n^{0}+3\times 4n^{3-1}+2\times 2n^{2-1}$$

Multiply 3 times 4.

$$5n^{0}+12n^{3-1}+2\times 2n^{2-1}$$

Subtract 1 from 3.

$$5n^{0}+12n^{2}+2\times 2n^{2-1}$$

Multiply 3 times 4.

$$5n^{0}+12n^{2}+4n^{2-1}$$

Subtract 1 from 2.

$$5n^{0}+12n^{2}+4n^{1}$$

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

$$5n^{0}+12n^{2}+4n$$

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

$$5\times 1+12n^{2}+4n$$

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

$$5+12n^{2}+4n$$