Solve 3x^3-2x^2-6x+4-7x^2-3

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

Evaluate

$$3x^{3}-9x^{2}-6x+1$$

Solution Steps

Combine -2x^{2} and -7x^{2} to get -9x^{2}.

$$3x^{3}-9x^{2}-6x+4-3$$

Subtract 3 from 4 to get 1.

$$3x^{3}-9x^{2}-6x+1$$

Differentiate w.r.t. x

$$9x^{2}-18x-6$$

Steps Using Derivative Rule for Sum

Combine -2x^{2} and -7x^{2} to get -9x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{3}-9x^{2}-6x+4-3)$$

Subtract 3 from 4 to get 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{3}-9x^{2}-6x+1)$$

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}.

$$3\times 3x^{3-1}+2\left(-9\right)x^{2-1}-6x^{1-1}$$

Multiply 3 times 3.

$$9x^{3-1}+2\left(-9\right)x^{2-1}-6x^{1-1}$$

Subtract 1 from 3.

$$9x^{2}+2\left(-9\right)x^{2-1}-6x^{1-1}$$

Multiply 2 times -9.

$$9x^{2}-18x^{2-1}-6x^{1-1}$$

Subtract 1 from 2.

$$9x^{2}-18x^{1}-6x^{1-1}$$

Subtract 1 from 1.

$$9x^{2}-18x^{1}-6x^{0}$$

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

$$9x^{2}-18x-6x^{0}$$

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

$$9x^{2}-18x-6$$