Solve (3a^2-3a+2)+(4a-a^2+5)=

$$( 3 a ^ { 2 } - 3 a + 2 ) + ( 4 a - a ^ { 2 } + 5 ) =$$

Evaluate

$$2a^{2}+a+7$$

Solution Steps

Combine -3a and 4a to get a.

$$3a^{2}+a+2-a^{2}+5$$

Combine 3a^{2} and -a^{2} to get 2a^{2}.

$$2a^{2}+a+2+5$$

Add 2 and 5 to get 7.

$$2a^{2}+a+7$$

Differentiate w.r.t. a

$$4a+1$$

Steps Using Derivative Rule for Sum

Combine -3a and 4a to get a.

$$\frac{\mathrm{d}}{\mathrm{d}a}(3a^{2}+a+2-a^{2}+5)$$

Combine 3a^{2} and -a^{2} to get 2a^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}a}(2a^{2}+a+2+5)$$

Add 2 and 5 to get 7.

$$\frac{\mathrm{d}}{\mathrm{d}a}(2a^{2}+a+7)$$

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}.

$$2\times 2a^{2-1}+a^{1-1}$$

Multiply 2 times 2.

$$4a^{2-1}+a^{1-1}$$

Subtract 1 from 2.

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

Subtract 1 from 1.

$$4a^{1}+a^{0}$$

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

$$4a+a^{0}$$

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

$$4a+1$$