Solve (2x^{4}+7+5x^2)+(8-8x^3+4x^4-3x^2)

$$( 2 x ^ { 4 } + 7 + 5 x ^ { 2 } ) + ( 8 - 8 x ^ { 3 } + 4 x ^ { 4 } - 3 x ^ { 2 } )$$

Evaluate

$$6x^{4}-8x^{3}+2x^{2}+15$$

Solution Steps

Add 7 and 8 to get 15.

$$2x^{4}+15+5x^{2}-8x^{3}+4x^{4}-3x^{2}$$

Combine 2x^{4} and 4x^{4} to get 6x^{4}.

$$6x^{4}+15+5x^{2}-8x^{3}-3x^{2}$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

$$6x^{4}+15+2x^{2}-8x^{3}$$

Differentiate w.r.t. x

$$4x\left(6x^{2}-6x+1\right)$$

Steps Using Derivative Rule for Sum

Add 7 and 8 to get 15.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{4}+15+5x^{2}-8x^{3}+4x^{4}-3x^{2})$$

Combine 2x^{4} and 4x^{4} to get 6x^{4}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(6x^{4}+15+5x^{2}-8x^{3}-3x^{2})$$

Combine 5x^{2} and -3x^{2} to get 2x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(6x^{4}+15+2x^{2}-8x^{3})$$

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}.

$$4\times 6x^{4-1}+2\times 2x^{2-1}+3\left(-8\right)x^{3-1}$$

Multiply 4 times 6.

$$24x^{4-1}+2\times 2x^{2-1}+3\left(-8\right)x^{3-1}$$

Subtract 1 from 4.

$$24x^{3}+2\times 2x^{2-1}+3\left(-8\right)x^{3-1}$$

Multiply 2 times 2.

$$24x^{3}+4x^{2-1}+3\left(-8\right)x^{3-1}$$

Subtract 1 from 2.

$$24x^{3}+4x^{1}+3\left(-8\right)x^{3-1}$$

Multiply 2 times 2.

$$24x^{3}+4x^{1}-24x^{3-1}$$

Subtract 1 from 3.

$$24x^{3}+4x^{1}-24x^{2}$$

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

$$24x^{3}+4x-24x^{2}$$