Solve 5-3x^3-3x+1+2x^3+7x-13

$$5 - 3 x ^ { 3 } - 3 x + 1 + 2 x ^ { 3 } + 7 x - 13$$

Evaluate

$$-x^{3}+4x-7$$

Solution Steps

Add 5 and 1 to get 6.

$$6-3x^{3}-3x+2x^{3}+7x-13$$

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

$$6-x^{3}-3x+7x-13$$

Combine -3x and 7x to get 4x.

$$6-x^{3}+4x-13$$

Subtract 13 from 6 to get -7.

$$-7-x^{3}+4x$$

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Sum

Add 5 and 1 to get 6.

$$\frac{\mathrm{d}}{\mathrm{d}x}(6-3x^{3}-3x+2x^{3}+7x-13)$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(6-x^{3}-3x+7x-13)$$

Combine -3x and 7x to get 4x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(6-x^{3}+4x-13)$$

Subtract 13 from 6 to get -7.

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

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}.

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

Multiply 3 times -1.

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

Subtract 1 from 3.

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

Subtract 1 from 1.

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

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

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

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

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