Solve 6^3+b^3+b^2+b

$$6 ^ { 3 } + b ^ { 3 } + b ^ { 2 } + b$$

Differentiate w.r.t. b

$$3b^{2}+2b+1$$

Steps Using Derivative Rule for Sum

Calculate 6 to the power of 3 and get 216.

$$\frac{\mathrm{d}}{\mathrm{d}b}(216+b^{3}+b^{2}+b)$$

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}.

$$3b^{3-1}+2b^{2-1}+b^{1-1}$$

Subtract 1 from 3.

$$3b^{2}+2b^{2-1}+b^{1-1}$$

Subtract 1 from 2.

$$3b^{2}+2b^{1}+b^{1-1}$$

Subtract 1 from 1.

$$3b^{2}+2b^{1}+b^{0}$$

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

$$3b^{2}+2b+b^{0}$$

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

$$3b^{2}+2b+1$$

Evaluate

$$b^{3}+b^{2}+b+216$$

Solution Steps

Calculate 6 to the power of 3 and get 216.

$$216+b^{3}+b^{2}+b$$