Solve =(3x^2+x-12)+(4x^3-2x^2+7x+10)

$$= ( 3 x ^ { 2 } + x - 12 ) + ( 4 x ^ { 3 } - 2 x ^ { 2 } + 7 x + 10 )$$

Evaluate

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

Solution Steps

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

$$x^{2}+x-12+4x^{3}+7x+10$$

Combine x and 7x to get 8x.

$$x^{2}+8x-12+4x^{3}+10$$

Add -12 and 10 to get -2.

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

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Sum

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+x-12+4x^{3}+7x+10)$$

Combine x and 7x to get 8x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{2}+8x-12+4x^{3}+10)$$

Add -12 and 10 to get -2.

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

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}.

$$2x^{2-1}+8x^{1-1}+3\times 4x^{3-1}$$

Subtract 1 from 2.

$$2x^{1}+8x^{1-1}+3\times 4x^{3-1}$$

Subtract 1 from 1.

$$2x^{1}+8x^{0}+3\times 4x^{3-1}$$

Multiply 1 times 8.

$$2x^{1}+8x^{0}+12x^{3-1}$$

Subtract 1 from 3.

$$2x^{1}+8x^{0}+12x^{2}$$

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

$$2x+8x^{0}+12x^{2}$$

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

$$2x+8\times 1+12x^{2}$$

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

$$2x+8+12x^{2}$$