Solve 4x+13-9x+1x^2+7x+1x^2

$$4 x + 13 - 9 x + 1 x ^ { 2 } + 7 x + 1 x ^ { 2 }$$

Evaluate

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

Solution Steps

Combine 4x and -9x to get -5x.

$$-5x+13+1x^{2}+7x+1x^{2}$$

Combine -5x and 7x to get 2x.

$$2x+13+1x^{2}+1x^{2}$$

Combine 1x^{2} and 1x^{2} to get 2x^{2}.

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

Differentiate w.r.t. x

$$4x+2$$

Steps Using Derivative Rule for Sum

Combine 4x and -9x to get -5x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-5x+13+1x^{2}+7x+1x^{2})$$

Combine -5x and 7x to get 2x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x+13+1x^{2}+1x^{2})$$

Combine 1x^{2} and 1x^{2} to get 2x^{2}.

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

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}.

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

Subtract 1 from 1.

$$2x^{0}+2\times 2x^{2-1}$$

Multiply 2 times 2.

$$2x^{0}+4x^{2-1}$$

Subtract 1 from 2.

$$2x^{0}+4x^{1}$$

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

$$2x^{0}+4x$$

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

$$2\times 1+4x$$

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

$$2+4x$$