Solve {l}{2n^3+4n-7}{+-n^3+8n}{-9}

$$\left. \begin{array} { l } { 2 n ^ { 3 } + 4 n - 7 } \\ { + - n ^ { 3 } + 8 n } \\ { - 9 } \end{array} \right.$$

Evaluate

$$n^{3}+12n-16$$

Solution Steps

Combine 4n and 8n to get 12n.

$$2n^{3}+12n-7-n^{3}-9$$

Subtract 9 from -7 to get -16.

$$2n^{3}+12n-16-n^{3}$$

Combine 2n^{3} and -n^{3} to get n^{3}.

$$n^{3}+12n-16$$

Differentiate w.r.t. n

$$3\left(n^{2}+4\right)$$

Steps Using Derivative Rule for Sum

Combine 4n and 8n to get 12n.

$$\frac{\mathrm{d}}{\mathrm{d}n}(2n^{3}+12n-7-n^{3}-9)$$

Subtract 9 from -7 to get -16.

$$\frac{\mathrm{d}}{\mathrm{d}n}(2n^{3}+12n-16-n^{3})$$

Combine 2n^{3} and -n^{3} to get n^{3}.

$$\frac{\mathrm{d}}{\mathrm{d}n}(n^{3}+12n-16)$$

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}.

$$3n^{3-1}+12n^{1-1}$$

Subtract 1 from 3.

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

Subtract 1 from 1.

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

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

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

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

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