Solve (4x^2-4x-9)+(-8x^2+7x-5)

$$( 4 x ^ { 2 } - 4 x - 9 ) + ( - 8 x ^ { 2 } + 7 x - 5 )$$

Evaluate

$$-4x^{2}+3x-14$$

Solution Steps

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$-4x^{2}-4x-9+7x-5$$

Combine -4x and 7x to get 3x.

$$-4x^{2}+3x-9-5$$

Subtract 5 from -9 to get -14.

$$-4x^{2}+3x-14$$

Differentiate w.r.t. x

$$3-8x$$

Steps Using Derivative Rule for Sum

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-4x^{2}-4x-9+7x-5)$$

Combine -4x and 7x to get 3x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-4x^{2}+3x-9-5)$$

Subtract 5 from -9 to get -14.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-4x^{2}+3x-14)$$

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}.

$$2\left(-4\right)x^{2-1}+3x^{1-1}$$

Multiply 2 times -4.

$$-8x^{2-1}+3x^{1-1}$$

Subtract 1 from 2.

$$-8x^{1}+3x^{1-1}$$

Subtract 1 from 1.

$$-8x^{1}+3x^{0}$$

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

$$-8x+3x^{0}$$

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

$$-8x+3\times 1$$

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

$$-8x+3$$