Solve (8x^2-5x+3)+(6x^3-2x-4)=

$$( 8 x ^ { 2 } - 5 x + 3 ) + ( 6 x ^ { 3 } - 2 x - 4 ) =$$

Evaluate

$$6x^{3}+8x^{2}-7x-1$$

Solution Steps

Combine -5x and -2x to get -7x.

$$8x^{2}-7x+3+6x^{3}-4$$

Subtract 4 from 3 to get -1.

$$8x^{2}-7x-1+6x^{3}$$

Differentiate w.r.t. x

$$18x^{2}+16x-7$$

Steps Using Derivative Rule for Sum

Combine -5x and -2x to get -7x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x^{2}-7x+3+6x^{3}-4)$$

Subtract 4 from 3 to get -1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x^{2}-7x-1+6x^{3})$$

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}.

$$2\times 8x^{2-1}-7x^{1-1}+3\times 6x^{3-1}$$

Multiply 2 times 8.

$$16x^{2-1}-7x^{1-1}+3\times 6x^{3-1}$$

Subtract 1 from 2.

$$16x^{1}-7x^{1-1}+3\times 6x^{3-1}$$

Subtract 1 from 1.

$$16x^{1}-7x^{0}+3\times 6x^{3-1}$$

Multiply 1 times -7.

$$16x^{1}-7x^{0}+18x^{3-1}$$

Subtract 1 from 3.

$$16x^{1}-7x^{0}+18x^{2}$$

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

$$16x-7x^{0}+18x^{2}$$

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

$$16x-7+18x^{2}$$