Solve -24+4(2)^3+k(2)^2-2(2)^2-1

$$- 24 + 4 ( 2 ) ^ { 3 } + k ( 2 ) ^ { 2 } - 2 ( 2 ) ^ { 2 } - 1$$

Evaluate

$$4k-1$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$-24+4\times 2^{3}+k\times 2^{2}-2^{3}-1$$

Calculate 2 to the power of 3 and get 8.

$$-24+4\times 8+k\times 2^{2}-2^{3}-1$$

Multiply 4 and 8 to get 32.

$$-24+32+k\times 2^{2}-2^{3}-1$$

Add -24 and 32 to get 8.

$$8+k\times 2^{2}-2^{3}-1$$

Calculate 2 to the power of 2 and get 4.

$$8+k\times 4-2^{3}-1$$

Calculate 2 to the power of 3 and get 8.

$$8+k\times 4-8-1$$

Subtract 8 from 8 to get 0.

$$k\times 4-1$$

Differentiate w.r.t. k

$$4$$

Steps Using Derivative Rule for Sum

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$\frac{\mathrm{d}}{\mathrm{d}k}(-24+4\times 2^{3}+k\times 2^{2}-2^{3}-1)$$

Calculate 2 to the power of 3 and get 8.

$$\frac{\mathrm{d}}{\mathrm{d}k}(-24+4\times 8+k\times 2^{2}-2^{3}-1)$$

Multiply 4 and 8 to get 32.

$$\frac{\mathrm{d}}{\mathrm{d}k}(-24+32+k\times 2^{2}-2^{3}-1)$$

Add -24 and 32 to get 8.

$$\frac{\mathrm{d}}{\mathrm{d}k}(8+k\times 2^{2}-2^{3}-1)$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\mathrm{d}}{\mathrm{d}k}(8+k\times 4-2^{3}-1)$$

Calculate 2 to the power of 3 and get 8.

$$\frac{\mathrm{d}}{\mathrm{d}k}(8+k\times 4-8-1)$$

Subtract 8 from 8 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}k}(k\times 4-1)$$

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}.

$$4k^{1-1}$$

Subtract 1 from 1.

$$4k^{0}$$

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

$$4\times 1$$

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

$$4$$