Solve (-x^2-5x+4)+(6x^2-8x+9)

$$( - x ^ { 2 } - 5 x + 4 ) + ( 6 x ^ { 2 } - 8 x + 9 )$$

Evaluate

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

Solution Steps

Combine -5x and -8x to get -13x.

$$-x^{2}-13x+4+6x^{2}+9$$

Add 4 and 9 to get 13.

$$-x^{2}-13x+13+6x^{2}$$

Combine -x^{2} and 6x^{2} to get 5x^{2}.

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

Differentiate w.r.t. x

$$10x-13$$

Steps Using Derivative Rule for Sum

Combine -5x and -8x to get -13x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-x^{2}-13x+4+6x^{2}+9)$$

Add 4 and 9 to get 13.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-x^{2}-13x+13+6x^{2})$$

Combine -x^{2} and 6x^{2} to get 5x^{2}.

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

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}.

$$2\times 5x^{2-1}-13x^{1-1}$$

Multiply 2 times 5.

$$10x^{2-1}-13x^{1-1}$$

Subtract 1 from 2.

$$10x^{1}-13x^{1-1}$$

Subtract 1 from 1.

$$10x^{1}-13x^{0}$$

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

$$10x-13x^{0}$$

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

$$10x-13$$