Solve (3y^3+6y-5)+(3y^2-5y+1)

$$( 3 y ^ { 3 } + 6 y - 5 ) + ( 3 y ^ { 2 } - 5 y + 1 )$$

Evaluate

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

Solution Steps

Combine 6y and -5y to get y.

$$3y^{3}+y-5+3y^{2}+1$$

Add -5 and 1 to get -4.

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

Differentiate w.r.t. y

$$\left(3y+1\right)^{2}$$

Steps Using Derivative Rule for Sum

Combine 6y and -5y to get y.

$$\frac{\mathrm{d}}{\mathrm{d}y}(3y^{3}+y-5+3y^{2}+1)$$

Add -5 and 1 to get -4.

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

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}.

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

Multiply 3 times 3.

$$9y^{3-1}+y^{1-1}+2\times 3y^{2-1}$$

Subtract 1 from 3.

$$9y^{2}+y^{1-1}+2\times 3y^{2-1}$$

Subtract 1 from 1.

$$9y^{2}+y^{0}+2\times 3y^{2-1}$$

Multiply 1 times 1.

$$9y^{2}+y^{0}+6y^{2-1}$$

Subtract 1 from 2.

$$9y^{2}+y^{0}+6y^{1}$$

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

$$9y^{2}+y^{0}+6y$$

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

$$9y^{2}+1+6y$$