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

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

Evaluate

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

Solution Steps

Combine 2x^{2} and x^{2} to get 3x^{2}.

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

Combine 2x and 3x to get 5x.

$$3x^{2}+5x-4+7$$

Add -4 and 7 to get 3.

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

Differentiate w.r.t. x

$$6x+5$$

Steps Using Derivative Rule for Sum

Combine 2x^{2} and x^{2} to get 3x^{2}.

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

Combine 2x and 3x to get 5x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}+5x-4+7)$$

Add -4 and 7 to get 3.

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

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}.

$$2\times 3x^{2-1}+5x^{1-1}$$

Multiply 2 times 3.

$$6x^{2-1}+5x^{1-1}$$

Subtract 1 from 2.

$$6x^{1}+5x^{1-1}$$

Subtract 1 from 1.

$$6x^{1}+5x^{0}$$

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

$$6x+5x^{0}$$

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

$$6x+5\times 1$$

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

$$6x+5$$