Solve 9(2)^3+k(2)+(k-1)

$$9 ( 2 ) ^ { 3 } + k ( 2 ) + ( k - 1 )$$

Evaluate

$$3k+71$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$9\times 8+k\times 2+k-1$$

Multiply 9 and 8 to get 72.

$$72+k\times 2+k-1$$

Combine k\times 2 and k to get 3k.

$$72+3k-1$$

Subtract 1 from 72 to get 71.

$$71+3k$$

Differentiate w.r.t. k

$$3$$

Steps Using Derivative Rule for Sum

Calculate 2 to the power of 3 and get 8.

$$\frac{\mathrm{d}}{\mathrm{d}k}(9\times 8+k\times 2+k-1)$$

Multiply 9 and 8 to get 72.

$$\frac{\mathrm{d}}{\mathrm{d}k}(72+k\times 2+k-1)$$

Combine k\times 2 and k to get 3k.

$$\frac{\mathrm{d}}{\mathrm{d}k}(72+3k-1)$$

Subtract 1 from 72 to get 71.

$$\frac{\mathrm{d}}{\mathrm{d}k}(71+3k)$$

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}.

$$3k^{1-1}$$

Subtract 1 from 1.

$$3k^{0}$$

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

$$3\times 1$$

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

$$3$$