Solve (7y^2+4y^3+2y)+(7+y+3y^2+y)

$$( 7 y ^ { 2 } + 4 y ^ { 3 } + 2 y ) + ( 7 + y + 3 y ^ { 2 } + y )$$

Evaluate

$$4y^{3}+10y^{2}+4y+7$$

Solution Steps

Combine 2y and y to get 3y.

$$7y^{2}+4y^{3}+3y+7+3y^{2}+y$$

Combine 7y^{2} and 3y^{2} to get 10y^{2}.

$$10y^{2}+4y^{3}+3y+7+y$$

Combine 3y and y to get 4y.

$$10y^{2}+4y^{3}+4y+7$$

Differentiate w.r.t. y

$$12y^{2}+20y+4$$

Steps Using Derivative Rule for Sum

Combine 2y and y to get 3y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(7y^{2}+4y^{3}+3y+7+3y^{2}+y)$$

Combine 7y^{2} and 3y^{2} to get 10y^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(10y^{2}+4y^{3}+3y+7+y)$$

Combine 3y and y to get 4y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(10y^{2}+4y^{3}+4y+7)$$

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}.

$$2\times 10y^{2-1}+3\times 4y^{3-1}+4y^{1-1}$$

Multiply 2 times 10.

$$20y^{2-1}+3\times 4y^{3-1}+4y^{1-1}$$

Subtract 1 from 2.

$$20y^{1}+3\times 4y^{3-1}+4y^{1-1}$$

Multiply 3 times 4.

$$20y^{1}+12y^{3-1}+4y^{1-1}$$

Subtract 1 from 3.

$$20y^{1}+12y^{2}+4y^{1-1}$$

Subtract 1 from 1.

$$20y^{1}+12y^{2}+4y^{0}$$

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

$$20y+12y^{2}+4y^{0}$$

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

$$20y+12y^{2}+4\times 1$$

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

$$20y+12y^{2}+4$$