Solve (-9x^2+5x-10)+(2x^2-2x+2)

$$(-9 { x }^{ 2 } +5x-10)+(2 { x }^{ 2 } -2x+2)$$

Evaluate

$$-7x^{2}+3x-8$$

Solution Steps

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

$$-7x^{2}+5x-10-2x+2$$

Combine 5x and -2x to get 3x.

$$-7x^{2}+3x-10+2$$

Add -10 and 2 to get -8.

$$-7x^{2}+3x-8$$

Differentiate w.r.t. x

$$3-14x$$

Steps Using Derivative Rule for Sum

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(-7x^{2}+5x-10-2x+2)$$

Combine 5x and -2x to get 3x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-7x^{2}+3x-10+2)$$

Add -10 and 2 to get -8.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-7x^{2}+3x-8)$$

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}.

$$2\left(-7\right)x^{2-1}+3x^{1-1}$$

Multiply 2 times -7.

$$-14x^{2-1}+3x^{1-1}$$

Subtract 1 from 2.

$$-14x^{1}+3x^{1-1}$$

Subtract 1 from 1.

$$-14x^{1}+3x^{0}$$

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

$$-14x+3x^{0}$$

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

$$-14x+3\times 1$$

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

$$-14x+3$$