Solve (6x^3+3x^2+3)+(2x^3-5x+1)

$$( 6 x ^ { 3 } + 3 x ^ { 2 } + 3 ) + ( 2 x ^ { 3 } - 5 x + 1 )$$

Evaluate

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

Solution Steps

Combine 6x^{3} and 2x^{3} to get 8x^{3}.

$$8x^{3}+3x^{2}+3-5x+1$$

Add 3 and 1 to get 4.

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

Differentiate w.r.t. x

$$24x^{2}+6x-5$$

Steps Using Derivative Rule for Sum

Combine 6x^{3} and 2x^{3} to get 8x^{3}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x^{3}+3x^{2}+3-5x+1)$$

Add 3 and 1 to get 4.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x^{3}+3x^{2}+4-5x)$$

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}.

$$3\times 8x^{3-1}+2\times 3x^{2-1}-5x^{1-1}$$

Multiply 3 times 8.

$$24x^{3-1}+2\times 3x^{2-1}-5x^{1-1}$$

Subtract 1 from 3.

$$24x^{2}+2\times 3x^{2-1}-5x^{1-1}$$

Multiply 2 times 3.

$$24x^{2}+6x^{2-1}-5x^{1-1}$$

Subtract 1 from 2.

$$24x^{2}+6x^{1}-5x^{1-1}$$

Subtract 1 from 1.

$$24x^{2}+6x^{1}-5x^{0}$$

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

$$24x^{2}+6x-5x^{0}$$

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

$$24x^{2}+6x-5$$