Solve (x^3+3x-5)+(2x^2-3x-5)+(x^2+4x-5)=

$$( x ^ { 3 } + 3 x - 5 ) + ( 2 x ^ { 2 } - 3 x - 5 ) + ( x ^ { 2 } + 4 x - 5 ) =$$

Evaluate

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

Solution Steps

Combine 3x and -3x to get 0.

$$x^{3}-5+2x^{2}-5+x^{2}+4x-5$$

Subtract 5 from -5 to get -10.

$$x^{3}-10+2x^{2}+x^{2}+4x-5$$

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

$$x^{3}-10+3x^{2}+4x-5$$

Subtract 5 from -10 to get -15.

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

Differentiate w.r.t. x

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

Steps Using Derivative Rule for Sum

Combine 3x and -3x to get 0.

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

Subtract 5 from -5 to get -10.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}-10+2x^{2}+x^{2}+4x-5)$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}-10+3x^{2}+4x-5)$$

Subtract 5 from -10 to get -15.

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

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}.

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

Subtract 1 from 3.

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

Multiply 2 times 3.

$$3x^{2}+6x^{2-1}+4x^{1-1}$$

Subtract 1 from 2.

$$3x^{2}+6x^{1}+4x^{1-1}$$

Subtract 1 from 1.

$$3x^{2}+6x^{1}+4x^{0}$$

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

$$3x^{2}+6x+4x^{0}$$

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

$$3x^{2}+6x+4\times 1$$

For any term t, t\times 1=t and 1t=t.

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