Solve (16x^3-12x+1)+(4x-4)

$$( 16 x ^ { 3 } - 12 x + 1 ) + ( 4 x - 4 )$$

Evaluate

$$16x^{3}-8x-3$$

Solution Steps

Combine -12x and 4x to get -8x.

$$16x^{3}-8x+1-4$$

Subtract 4 from 1 to get -3.

$$16x^{3}-8x-3$$

Differentiate w.r.t. x

$$48x^{2}-8$$

Steps Using Derivative Rule for Sum

Combine -12x and 4x to get -8x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(16x^{3}-8x+1-4)$$

Subtract 4 from 1 to get -3.

$$\frac{\mathrm{d}}{\mathrm{d}x}(16x^{3}-8x-3)$$

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}.

$$3\times 16x^{3-1}-8x^{1-1}$$

Multiply 3 times 16.

$$48x^{3-1}-8x^{1-1}$$

Subtract 1 from 3.

$$48x^{2}-8x^{1-1}$$

Subtract 1 from 1.

$$48x^{2}-8x^{0}$$

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

$$48x^{2}-8$$