Solve +0x^{3}-8x^2-6x+3-2x^4-0x^3-0x^2+2x+2

$$+ 0 x ^ { 3 } - 8 x ^ { 2 } - 6 x + 3 - 2 x ^ { 4 } - 0 x ^ { 3 } - 0 x ^ { 2 } + 2 x + 2$$

Evaluate

$$5-4x-8x^{2}-2x^{4}$$

Solution Steps

Anything times zero gives zero.

$$0-8x^{2}-6x+3-2x^{4}-0x^{3}-0x^{2}+2x+2$$

Add 0 and 3 to get 3.

$$3-8x^{2}-6x-2x^{4}-0x^{3}-0x^{2}+2x+2$$

Anything times zero gives zero.

$$3-8x^{2}-6x-2x^{4}-0-0x^{2}+2x+2$$

Anything times zero gives zero.

$$3-8x^{2}-6x-2x^{4}-0-0+2x+2$$

Multiply -1 and 0 to get 0.

$$3-8x^{2}-6x-2x^{4}+0-0+2x+2$$

Add 3 and 0 to get 3.

$$3-8x^{2}-6x-2x^{4}-0+2x+2$$

Multiply -1 and 0 to get 0.

$$3-8x^{2}-6x-2x^{4}+0+2x+2$$

Add 3 and 0 to get 3.

$$3-8x^{2}-6x-2x^{4}+2x+2$$

Combine -6x and 2x to get -4x.

$$3-8x^{2}-4x-2x^{4}+2$$

Add 3 and 2 to get 5.

$$5-8x^{2}-4x-2x^{4}$$

Differentiate w.r.t. x

$$-8x^{3}-16x-4$$

Steps Using Derivative Rule for Sum

Anything times zero gives zero.

$$\frac{\mathrm{d}}{\mathrm{d}x}(0-8x^{2}-6x+3-2x^{4}-0x^{3}-0x^{2}+2x+2)$$

Add 0 and 3 to get 3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3-8x^{2}-6x-2x^{4}-0x^{3}-0x^{2}+2x+2)$$

Anything times zero gives zero.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3-8x^{2}-6x-2x^{4}-0-0x^{2}+2x+2)$$

Anything times zero gives zero.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3-8x^{2}-6x-2x^{4}-0-0+2x+2)$$

Multiply -1 and 0 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3-8x^{2}-6x-2x^{4}+0-0+2x+2)$$

Add 3 and 0 to get 3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3-8x^{2}-6x-2x^{4}-0+2x+2)$$

Multiply -1 and 0 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3-8x^{2}-6x-2x^{4}+0+2x+2)$$

Add 3 and 0 to get 3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3-8x^{2}-6x-2x^{4}+2x+2)$$

Combine -6x and 2x to get -4x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3-8x^{2}-4x-2x^{4}+2)$$

Add 3 and 2 to get 5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(5-8x^{2}-4x-2x^{4})$$

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}.

$$2\left(-8\right)x^{2-1}-4x^{1-1}+4\left(-2\right)x^{4-1}$$

Multiply 2 times -8.

$$-16x^{2-1}-4x^{1-1}+4\left(-2\right)x^{4-1}$$

Subtract 1 from 2.

$$-16x^{1}-4x^{1-1}+4\left(-2\right)x^{4-1}$$

Subtract 1 from 1.

$$-16x^{1}-4x^{0}+4\left(-2\right)x^{4-1}$$

Multiply 1 times -4.

$$-16x^{1}-4x^{0}-8x^{4-1}$$

Subtract 1 from 4.

$$-16x^{1}-4x^{0}-8x^{3}$$

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

$$-16x-4x^{0}-8x^{3}$$

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

$$-16x-4-8x^{3}$$