Solve (x^3-2x+3)+(3x^2+6x-2)

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

Evaluate

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

Solution Steps

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

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

Subtract 2 from 3 to get 1.

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

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Sum

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

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

Subtract 2 from 3 to get 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}+4x+1+3x^{2})$$

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}.

$$3x^{3-1}+4x^{1-1}+2\times 3x^{2-1}$$

Subtract 1 from 3.

$$3x^{2}+4x^{1-1}+2\times 3x^{2-1}$$

Subtract 1 from 1.

$$3x^{2}+4x^{0}+2\times 3x^{2-1}$$

Multiply 1 times 4.

$$3x^{2}+4x^{0}+6x^{2-1}$$

Subtract 1 from 2.

$$3x^{2}+4x^{0}+6x^{1}$$

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

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

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

$$3x^{2}+4\times 1+6x$$

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

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