Solve 23^3-k3^2+3(3)+10

$$2 { 3 }^{ 3 } -k { 3 }^{ 2 } +3(3)+10$$

Evaluate

$$73-9k$$

Solution Steps

Calculate 3 to the power of 3 and get 27.

$$2\times 27-k\times 3^{2}+3\times 3+10$$

Multiply 2 and 27 to get 54.

$$54-k\times 3^{2}+3\times 3+10$$

Calculate 3 to the power of 2 and get 9.

$$54-k\times 9+3\times 3+10$$

Multiply 3 and 3 to get 9.

$$54-k\times 9+9+10$$

Add 9 and 10 to get 19.

$$54-k\times 9+19$$

Multiply -1 and 9 to get -9.

$$54-9k+19$$

Add 54 and 19 to get 73.

$$73-9k$$

Differentiate w.r.t. k

$$-9$$

Steps Using Derivative Rule for Sum

Calculate 3 to the power of 3 and get 27.

$$\frac{\mathrm{d}}{\mathrm{d}k}(2\times 27-k\times 3^{2}+3\times 3+10)$$

Multiply 2 and 27 to get 54.

$$\frac{\mathrm{d}}{\mathrm{d}k}(54-k\times 3^{2}+3\times 3+10)$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\mathrm{d}}{\mathrm{d}k}(54-k\times 9+3\times 3+10)$$

Multiply 3 and 3 to get 9.

$$\frac{\mathrm{d}}{\mathrm{d}k}(54-k\times 9+9+10)$$

Add 9 and 10 to get 19.

$$\frac{\mathrm{d}}{\mathrm{d}k}(54-k\times 9+19)$$

Multiply -1 and 9 to get -9.

$$\frac{\mathrm{d}}{\mathrm{d}k}(54-9k+19)$$

Add 54 and 19 to get 73.

$$\frac{\mathrm{d}}{\mathrm{d}k}(73-9k)$$

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}.

$$-9k^{1-1}$$

Subtract 1 from 1.

$$-9k^{0}$$

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

$$-9$$