Solve 2x-x+3x+4x-3+4-5+6+1

$$2x-x+3x+4x-3+4-5+6+1$$

Evaluate

$$8x+3$$

Solution Steps

Combine 2x and -x to get x.

$$x+3x+4x-3+4-5+6+1$$

Combine x and 3x to get 4x.

$$4x+4x-3+4-5+6+1$$

Combine 4x and 4x to get 8x.

$$8x-3+4-5+6+1$$

Add -3 and 4 to get 1.

$$8x+1-5+6+1$$

Subtract 5 from 1 to get -4.

$$8x-4+6+1$$

Add -4 and 6 to get 2.

$$8x+2+1$$

Add 2 and 1 to get 3.

$$8x+3$$

Differentiate w.r.t. x

$$8$$

Steps Using Derivative Rule for Sum

Combine 2x and -x to get x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x+3x+4x-3+4-5+6+1)$$

Combine x and 3x to get 4x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(4x+4x-3+4-5+6+1)$$

Combine 4x and 4x to get 8x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x-3+4-5+6+1)$$

Add -3 and 4 to get 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x+1-5+6+1)$$

Subtract 5 from 1 to get -4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x-4+6+1)$$

Add -4 and 6 to get 2.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x+2+1)$$

Add 2 and 1 to get 3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x+3)$$

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}.

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

Subtract 1 from 1.

$$8x^{0}$$

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

$$8\times 1$$

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

$$8$$