Solve -2x^3-8x^2+6+3x-4x^2+3+8x^2+5

$$- 2 x ^ { 3 } - 8 x ^ { 2 } + 6 + 3 x - 4 x ^ { 2 } + 3 + 8 x ^ { 2 } + 5$$

Evaluate

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

Solution Steps

Combine -8x^{2} and -4x^{2} to get -12x^{2}.

$$-2x^{3}-12x^{2}+6+3x+3+8x^{2}+5$$

Add 6 and 3 to get 9.

$$-2x^{3}-12x^{2}+9+3x+8x^{2}+5$$

Combine -12x^{2} and 8x^{2} to get -4x^{2}.

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

Add 9 and 5 to get 14.

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

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Sum

Combine -8x^{2} and -4x^{2} to get -12x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-2x^{3}-12x^{2}+6+3x+3+8x^{2}+5)$$

Add 6 and 3 to get 9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-2x^{3}-12x^{2}+9+3x+8x^{2}+5)$$

Combine -12x^{2} and 8x^{2} to get -4x^{2}.

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

Add 9 and 5 to get 14.

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

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}.

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

Multiply 3 times -2.

$$-6x^{3-1}+2\left(-4\right)x^{2-1}+3x^{1-1}$$

Subtract 1 from 3.

$$-6x^{2}+2\left(-4\right)x^{2-1}+3x^{1-1}$$

Multiply 2 times -4.

$$-6x^{2}-8x^{2-1}+3x^{1-1}$$

Subtract 1 from 2.

$$-6x^{2}-8x^{1}+3x^{1-1}$$

Subtract 1 from 1.

$$-6x^{2}-8x^{1}+3x^{0}$$

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

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

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

$$-6x^{2}-8x+3\times 1$$

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

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