Solve (2t^3-3t^2+1)+(4t^3+5t^2+2t)

$$( 2 t ^ { 3 } - 3 t ^ { 2 } + 1 ) + ( 4 t ^ { 3 } + 5 t ^ { 2 } + 2 t )$$

Evaluate

$$6t^{3}+2t^{2}+2t+1$$

Solution Steps

Combine 2t^{3} and 4t^{3} to get 6t^{3}.

$$6t^{3}-3t^{2}+1+5t^{2}+2t$$

Combine -3t^{2} and 5t^{2} to get 2t^{2}.

$$6t^{3}+2t^{2}+1+2t$$

Differentiate w.r.t. t

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

Steps Using Derivative Rule for Sum

Combine 2t^{3} and 4t^{3} to get 6t^{3}.

$$\frac{\mathrm{d}}{\mathrm{d}t}(6t^{3}-3t^{2}+1+5t^{2}+2t)$$

Combine -3t^{2} and 5t^{2} to get 2t^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}t}(6t^{3}+2t^{2}+1+2t)$$

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}.

$$3\times 6t^{3-1}+2\times 2t^{2-1}+2t^{1-1}$$

Multiply 3 times 6.

$$18t^{3-1}+2\times 2t^{2-1}+2t^{1-1}$$

Subtract 1 from 3.

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

Multiply 2 times 2.

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

Subtract 1 from 2.

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

Subtract 1 from 1.

$$18t^{2}+4t^{1}+2t^{0}$$

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

$$18t^{2}+4t+2t^{0}$$

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

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

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

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