Solve x^2+3x+x+3-2x^2-3x+3+2x^2-6x-x+4

$${ x }^{ 2 } +3x+x+3-2 { x }^{ 2 } -3x+3+2 { x }^{ 2 } -6x-x+4$$

Evaluate

$$x^{2}-6x+10$$

Solution Steps

Combine 3x and x to get 4x.

$$x^{2}+4x+3-2x^{2}-3x+3+2x^{2}-6x-x+4$$

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

$$-x^{2}+4x+3-3x+3+2x^{2}-6x-x+4$$

Combine 4x and -3x to get x.

$$-x^{2}+x+3+3+2x^{2}-6x-x+4$$

Add 3 and 3 to get 6.

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

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

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

Combine x and -6x to get -5x.

$$x^{2}-5x+6-x+4$$

Combine -5x and -x to get -6x.

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

Add 6 and 4 to get 10.

$$x^{2}-6x+10$$

Differentiate w.r.t. x

$$2\left(x-3\right)$$

Steps Using Derivative Rule for Sum

Combine 3x and x to get 4x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+4x+3-2x^{2}-3x+3+2x^{2}-6x-x+4)$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(-x^{2}+4x+3-3x+3+2x^{2}-6x-x+4)$$

Combine 4x and -3x to get x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-x^{2}+x+3+3+2x^{2}-6x-x+4)$$

Add 3 and 3 to get 6.

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

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

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

Combine x and -6x to get -5x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-5x+6-x+4)$$

Combine -5x and -x to get -6x.

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

Add 6 and 4 to get 10.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-6x+10)$$

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}.

$$2x^{2-1}-6x^{1-1}$$

Subtract 1 from 2.

$$2x^{1}-6x^{1-1}$$

Subtract 1 from 1.

$$2x^{1}-6x^{0}$$

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

$$2x-6x^{0}$$

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

$$2x-6$$