Solve 4x^2-1+x^3-3x^2+6x-2

$$4 x ^ { 2 } - 1 + x ^ { 3 } - 3 x ^ { 2 } + 6 x - 2$$

Evaluate

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

Solution Steps

Combine 4x^{2} and -3x^{2} to get x^{2}.

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

Subtract 2 from -1 to get -3.

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

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Sum

Combine 4x^{2} and -3x^{2} to get x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}-1+x^{3}+6x-2)$$

Subtract 2 from -1 to get -3.

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

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}.

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

Subtract 1 from 2.

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

Subtract 1 from 3.

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

Subtract 1 from 1.

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

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

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

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

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

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

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