Solve (-7x^2-3x+4)+(9x^2+6x+7)

$$( - 7 x ^ { 2 } - 3 x + 4 ) + ( 9 x ^ { 2 } + 6 x + 7 )$$

Evaluate

$$2x^{2}+3x+11$$

Solution Steps

Combine -7x^{2} and 9x^{2} to get 2x^{2}.

$$2x^{2}-3x+4+6x+7$$

Combine -3x and 6x to get 3x.

$$2x^{2}+3x+4+7$$

Add 4 and 7 to get 11.

$$2x^{2}+3x+11$$

Differentiate w.r.t. x

$$4x+3$$

Steps Using Derivative Rule for Sum

Combine -7x^{2} and 9x^{2} to get 2x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}-3x+4+6x+7)$$

Combine -3x and 6x to get 3x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}+3x+4+7)$$

Add 4 and 7 to get 11.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}+3x+11)$$

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}.

$$2\times 2x^{2-1}+3x^{1-1}$$

Multiply 2 times 2.

$$4x^{2-1}+3x^{1-1}$$

Subtract 1 from 2.

$$4x^{1}+3x^{1-1}$$

Subtract 1 from 1.

$$4x^{1}+3x^{0}$$

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

$$4x+3x^{0}$$

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

$$4x+3\times 1$$

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

$$4x+3$$