Solve (2x^2-3x+1-x^3)+1x+27=

$$( 2 x ^ { 2 } - 3 x + 1 - x ^ { 3 } ) + 1 x + 27 =$$

Evaluate

$$28-2x+2x^{2}-x^{3}$$

Solution Steps

Combine -3x and 1x to get -2x.

$$2x^{2}-2x+1-x^{3}+27$$

Add 1 and 27 to get 28.

$$2x^{2}-2x+28-x^{3}$$

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Sum

Combine -3x and 1x to get -2x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}-2x+1-x^{3}+27)$$

Add 1 and 27 to get 28.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}-2x+28-x^{3})$$

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}.

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

Multiply 2 times 2.

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

Subtract 1 from 2.

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

Subtract 1 from 1.

$$4x^{1}-2x^{0}+3\left(-1\right)x^{3-1}$$

Multiply 1 times -2.

$$4x^{1}-2x^{0}-3x^{3-1}$$

Subtract 1 from 3.

$$4x^{1}-2x^{0}-3x^{2}$$

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

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

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

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