Solve 4x^3-2x^2-28x-6x-5x-15

$$4 x ^ { 3 } - 2 x ^ { 2 } - 28 x - 6 x - 5 x - 15$$

Evaluate

$$4x^{3}-2x^{2}-39x-15$$

Solution Steps

Combine -28x and -6x to get -34x.

$$4x^{3}-2x^{2}-34x-5x-15$$

Combine -34x and -5x to get -39x.

$$4x^{3}-2x^{2}-39x-15$$

Differentiate w.r.t. x

$$12x^{2}-4x-39$$

Steps Using Derivative Rule for Sum

Combine -28x and -6x to get -34x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(4x^{3}-2x^{2}-34x-5x-15)$$

Combine -34x and -5x to get -39x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(4x^{3}-2x^{2}-39x-15)$$

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}.

$$3\times 4x^{3-1}+2\left(-2\right)x^{2-1}-39x^{1-1}$$

Multiply 3 times 4.

$$12x^{3-1}+2\left(-2\right)x^{2-1}-39x^{1-1}$$

Subtract 1 from 3.

$$12x^{2}+2\left(-2\right)x^{2-1}-39x^{1-1}$$

Multiply 2 times -2.

$$12x^{2}-4x^{2-1}-39x^{1-1}$$

Subtract 1 from 2.

$$12x^{2}-4x^{1}-39x^{1-1}$$

Subtract 1 from 1.

$$12x^{2}-4x^{1}-39x^{0}$$

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

$$12x^{2}-4x-39x^{0}$$

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

$$12x^{2}-4x-39$$