Solve (9x^2+3x+5)+(x^2+5x+8)

$$( 9 x ^ { 2 } + 3 x + 5 ) + ( x ^ { 2 } + 5 x + 8 )$$

Evaluate

$$10x^{2}+8x+13$$

Solution Steps

Combine 9x^{2} and x^{2} to get 10x^{2}.

$$10x^{2}+3x+5+5x+8$$

Combine 3x and 5x to get 8x.

$$10x^{2}+8x+5+8$$

Add 5 and 8 to get 13.

$$10x^{2}+8x+13$$

Differentiate w.r.t. x

$$20x+8$$

Steps Using Derivative Rule for Sum

Combine 9x^{2} and x^{2} to get 10x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(10x^{2}+3x+5+5x+8)$$

Combine 3x and 5x to get 8x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(10x^{2}+8x+5+8)$$

Add 5 and 8 to get 13.

$$\frac{\mathrm{d}}{\mathrm{d}x}(10x^{2}+8x+13)$$

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}.

$$2\times 10x^{2-1}+8x^{1-1}$$

Multiply 2 times 10.

$$20x^{2-1}+8x^{1-1}$$

Subtract 1 from 2.

$$20x^{1}+8x^{1-1}$$

Subtract 1 from 1.

$$20x^{1}+8x^{0}$$

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

$$20x+8x^{0}$$

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

$$20x+8\times 1$$

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

$$20x+8$$