Solve =10+20x-2x^2+10x^2-40x+5

$$= 10 + 20 x - 2 x ^ { 2 } + 10 x ^ { 2 } - 40 x + 5$$

Evaluate

$$8x^{2}-20x+15$$

Solution Steps

Combine -2x^{2} and 10x^{2} to get 8x^{2}.

$$10+20x+8x^{2}-40x+5$$

Combine 20x and -40x to get -20x.

$$10-20x+8x^{2}+5$$

Add 10 and 5 to get 15.

$$15-20x+8x^{2}$$

Differentiate w.r.t. x

$$16x-20$$

Steps Using Derivative Rule for Sum

Combine -2x^{2} and 10x^{2} to get 8x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(10+20x+8x^{2}-40x+5)$$

Combine 20x and -40x to get -20x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(10-20x+8x^{2}+5)$$

Add 10 and 5 to get 15.

$$\frac{\mathrm{d}}{\mathrm{d}x}(15-20x+8x^{2})$$

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}.

$$-20x^{1-1}+2\times 8x^{2-1}$$

Subtract 1 from 1.

$$-20x^{0}+2\times 8x^{2-1}$$

Multiply 2 times 8.

$$-20x^{0}+16x^{2-1}$$

Subtract 1 from 2.

$$-20x^{0}+16x^{1}$$

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

$$-20x^{0}+16x$$

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

$$-20+16x$$