Solve (2c+9)+(9c+8)

$$( 2 c + 9 ) + ( 9 c + 8 )$$

Evaluate

$$11c+17$$

Solution Steps

Combine 2c and 9c to get 11c.

$$11c+9+8$$

Add 9 and 8 to get 17.

$$11c+17$$

Differentiate w.r.t. c

$$11$$

Steps Using Derivative Rule for Sum

Combine 2c and 9c to get 11c.

$$\frac{\mathrm{d}}{\mathrm{d}c}(11c+9+8)$$

Add 9 and 8 to get 17.

$$\frac{\mathrm{d}}{\mathrm{d}c}(11c+17)$$

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}.

$$11c^{1-1}$$

Subtract 1 from 1.

$$11c^{0}$$

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

$$11\times 1$$

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

$$11$$