Solve (3x^3+3x^2-4x+5)+(x^3-2x^2+x-4)

$$( 3 x ^ { 3 } + 3 x ^ { 2 } - 4 x + 5 ) + ( x ^ { 3 } - 2 x ^ { 2 } + x - 4 )$$

Evaluate

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

Solution Steps

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

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

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

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

Combine -4x and x to get -3x.

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

Subtract 4 from 5 to get 1.

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

Differentiate w.r.t. x

$$12x^{2}+2x-3$$

Steps Using Derivative Rule for Sum

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

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

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

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

Combine -4x and x to get -3x.

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

Subtract 4 from 5 to get 1.

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

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}.

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

Multiply 3 times 4.

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

Subtract 1 from 3.

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

Subtract 1 from 2.

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

Subtract 1 from 1.

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

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

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

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

$$12x^{2}+2x-3$$