Solve (a-b+2c)+(3a+b-2c)=

$$( a - b + 2 c ) + ( 3 a + b - 2 c ) =$$

Evaluate

$$4a$$

Solution Steps

Combine a and 3a to get 4a.

$$4a-b+2c+b-2c$$

Combine -b and b to get 0.

$$4a+2c-2c$$

Combine 2c and -2c to get 0.

$$4a$$

Differentiate w.r.t. a

$$4$$

Steps Using Definition of a Derivative

Combine a and 3a to get 4a.

$$\frac{\mathrm{d}}{\mathrm{d}a}(4a-b+2c+b-2c)$$

Combine -b and b to get 0.

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

Combine 2c and -2c to get 0.

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

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

$$4a^{1-1}$$

Subtract 1 from 1.

$$4a^{0}$$

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

$$4\times 1$$

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

$$4$$