Solve (9t^2+4t+4)+(9t^2+3)

$$( 9 t ^ { 2 } + 4 t + 4 ) + ( 9 t ^ { 2 } + 3 )$$

Evaluate

$$18t^{2}+4t+7$$

Solution Steps

Combine 9t^{2} and 9t^{2} to get 18t^{2}.

$$18t^{2}+4t+4+3$$

Add 4 and 3 to get 7.

$$18t^{2}+4t+7$$

Differentiate w.r.t. t

$$36t+4$$

Steps Using Derivative Rule for Sum

Combine 9t^{2} and 9t^{2} to get 18t^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}t}(18t^{2}+4t+4+3)$$

Add 4 and 3 to get 7.

$$\frac{\mathrm{d}}{\mathrm{d}t}(18t^{2}+4t+7)$$

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}.

$$2\times 18t^{2-1}+4t^{1-1}$$

Multiply 2 times 18.

$$36t^{2-1}+4t^{1-1}$$

Subtract 1 from 2.

$$36t^{1}+4t^{1-1}$$

Subtract 1 from 1.

$$36t^{1}+4t^{0}$$

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

$$36t+4t^{0}$$

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

$$36t+4\times 1$$

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

$$36t+4$$