Solve (7y^2+y-8)+(9y^2-y-5)

$$( 7 y ^ { 2 } + y - 8 ) + ( 9 y ^ { 2 } - y - 5 )$$

Evaluate

$$16y^{2}-13$$

Solution Steps

Combine 7y^{2} and 9y^{2} to get 16y^{2}.

$$16y^{2}+y-8-y-5$$

Combine y and -y to get 0.

$$16y^{2}-8-5$$

Subtract 5 from -8 to get -13.

$$16y^{2}-13$$

Differentiate w.r.t. y

$$32y$$

Steps Using Derivative Rule for Sum

Combine 7y^{2} and 9y^{2} to get 16y^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}y}(16y^{2}+y-8-y-5)$$

Combine y and -y to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}y}(16y^{2}-8-5)$$

Subtract 5 from -8 to get -13.

$$\frac{\mathrm{d}}{\mathrm{d}y}(16y^{2}-13)$$

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}.

$$2\times 16y^{2-1}$$

Multiply 2 times 16.

$$32y^{2-1}$$

Subtract 1 from 2.

$$32y^{1}$$

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

$$32y$$