Solve {r}{x^2+6x+9}{+quadx^2+4x+4}

$$\left. \begin{array} { r } { x ^ { 2 } + 6 x + 9 } \\ { + \quad x ^ { 2 } + 4 x + 4 } \end{array} \right.$$

Evaluate

$$2x^{2}+10x+13$$

Solution Steps

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+6x+9+4x+4$$

Combine 6x and 4x to get 10x.

$$2x^{2}+10x+9+4$$

Add 9 and 4 to get 13.

$$2x^{2}+10x+13$$

Differentiate w.r.t. x

$$4x+10$$

Steps Using Derivative Rule for Sum

Combine x^{2} and x^{2} to get 2x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}+6x+9+4x+4)$$

Combine 6x and 4x to get 10x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}+10x+9+4)$$

Add 9 and 4 to get 13.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}+10x+13)$$

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}.

$$2\times 2x^{2-1}+10x^{1-1}$$

Multiply 2 times 2.

$$4x^{2-1}+10x^{1-1}$$

Subtract 1 from 2.

$$4x^{1}+10x^{1-1}$$

Subtract 1 from 1.

$$4x^{1}+10x^{0}$$

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

$$4x+10x^{0}$$

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

$$4x+10\times 1$$

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

$$4x+10$$