Solve (+2)(-(-6)c+(-8)c+(+27)c-(+5)c

$$( + 2 ) ( - ( - 6 ) c + ( - 8 ) c + ( + 27 ) c - ( + 5 ) c$$

Evaluate

$$40c$$

Solution Steps

The opposite of -6 is 6.

$$2\left(6c-8c+27c-5c\right)$$

Combine 6c and -8c to get -2c.

$$2\left(-2c+27c-5c\right)$$

Combine -2c and 27c to get 25c.

$$2\left(25c-5c\right)$$

Combine 25c and -5c to get 20c.

$$2\times 20c$$

Multiply 2 and 20 to get 40.

$$40c$$

Differentiate w.r.t. c

$$40$$

Steps Using Definition of a Derivative

The opposite of -6 is 6.

$$\frac{\mathrm{d}}{\mathrm{d}c}(2\left(6c-8c+27c-5c\right))$$

Combine 6c and -8c to get -2c.

$$\frac{\mathrm{d}}{\mathrm{d}c}(2\left(-2c+27c-5c\right))$$

Combine -2c and 27c to get 25c.

$$\frac{\mathrm{d}}{\mathrm{d}c}(2\left(25c-5c\right))$$

Combine 25c and -5c to get 20c.

$$\frac{\mathrm{d}}{\mathrm{d}c}(2\times 20c)$$

Multiply 2 and 20 to get 40.

$$\frac{\mathrm{d}}{\mathrm{d}c}(40c)$$

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

$$40c^{1-1}$$

Subtract 1 from 1.

$$40c^{0}$$

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

$$40\times 1$$

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

$$40$$