Solve (3x^3+6x-4)+(3x^3-4x^2+5x)

$$( 3 x ^ { 3 } + 6 x - 4 ) + ( 3 x ^ { 3 } - 4 x ^ { 2 } + 5 x )$$

Evaluate

$$6x^{3}-4x^{2}+11x-4$$

Solution Steps

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

$$6x^{3}+6x-4-4x^{2}+5x$$

Combine 6x and 5x to get 11x.

$$6x^{3}+11x-4-4x^{2}$$

Differentiate w.r.t. x

$$18x^{2}-8x+11$$

Steps Using Derivative Rule for Sum

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(6x^{3}+6x-4-4x^{2}+5x)$$

Combine 6x and 5x to get 11x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(6x^{3}+11x-4-4x^{2})$$

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}.

$$3\times 6x^{3-1}+11x^{1-1}+2\left(-4\right)x^{2-1}$$

Multiply 3 times 6.

$$18x^{3-1}+11x^{1-1}+2\left(-4\right)x^{2-1}$$

Subtract 1 from 3.

$$18x^{2}+11x^{1-1}+2\left(-4\right)x^{2-1}$$

Subtract 1 from 1.

$$18x^{2}+11x^{0}+2\left(-4\right)x^{2-1}$$

Multiply 1 times 11.

$$18x^{2}+11x^{0}-8x^{2-1}$$

Subtract 1 from 2.

$$18x^{2}+11x^{0}-8x^{1}$$

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

$$18x^{2}+11x^{0}-8x$$

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

$$18x^{2}+11\times 1-8x$$

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

$$18x^{2}+11-8x$$