Solve (2x^2-3x-1)+(x^2-x+4)

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

Evaluate

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

Solution Steps

Combine 2x^{2} and x^{2} to get 3x^{2}.

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

Combine -3x and -x to get -4x.

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

Add -1 and 4 to get 3.

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

Differentiate w.r.t. x

$$6x-4$$

Steps Using Derivative Rule for Sum

Combine 2x^{2} and x^{2} to get 3x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}-3x-1-x+4)$$

Combine -3x and -x to get -4x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}-4x-1+4)$$

Add -1 and 4 to get 3.

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

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}.

$$2\times 3x^{2-1}-4x^{1-1}$$

Multiply 2 times 3.

$$6x^{2-1}-4x^{1-1}$$

Subtract 1 from 2.

$$6x^{1}-4x^{1-1}$$

Subtract 1 from 1.

$$6x^{1}-4x^{0}$$

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

$$6x-4x^{0}$$

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

$$6x-4$$