Solve [(y+4)+(y-5)]+[(y+3)+(y-2)]=

$$[ ( y + 4 ) + ( y - 5 ) ] + [ ( y + 3 ) + ( y - 2 ) ] =$$

Evaluate

$$4y$$

Solution Steps

Combine y and y to get 2y.

$$2y+4-5+y+3+y-2$$

Subtract 5 from 4 to get -1.

$$2y-1+y+3+y-2$$

Combine 2y and y to get 3y.

$$3y-1+3+y-2$$

Add -1 and 3 to get 2.

$$3y+2+y-2$$

Combine 3y and y to get 4y.

$$4y+2-2$$

Subtract 2 from 2 to get 0.

$$4y$$

Differentiate w.r.t. y

$$4$$

Steps Using Definition of a Derivative

Combine y and y to get 2y.

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

Subtract 5 from 4 to get -1.

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

Combine 2y and y to get 3y.

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

Add -1 and 3 to get 2.

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

Combine 3y and y to get 4y.

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

Subtract 2 from 2 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}y}(4y)$$

The derivative of ax^{n} is nax^{n-1}.

$$4y^{1-1}$$

Subtract 1 from 1.

$$4y^{0}$$

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

$$4\times 1$$

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

$$4$$