Solve (-2)^4+4(2)^3-2(2)^2+n(2)^2-1

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

Evaluate

$$4n+39$$

Solution Steps

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

$$\left(-2\right)^{4}+4\times 2^{3}-2^{3}+n\times 2^{2}-1$$

Calculate -2 to the power of 4 and get 16.

$$16+4\times 2^{3}-2^{3}+n\times 2^{2}-1$$

Calculate 2 to the power of 3 and get 8.

$$16+4\times 8-2^{3}+n\times 2^{2}-1$$

Multiply 4 and 8 to get 32.

$$16+32-2^{3}+n\times 2^{2}-1$$

Add 16 and 32 to get 48.

$$48-2^{3}+n\times 2^{2}-1$$

Calculate 2 to the power of 3 and get 8.

$$48-8+n\times 2^{2}-1$$

Subtract 8 from 48 to get 40.

$$40+n\times 2^{2}-1$$

Calculate 2 to the power of 2 and get 4.

$$40+n\times 4-1$$

Subtract 1 from 40 to get 39.

$$39+n\times 4$$

Differentiate w.r.t. n

$$4$$

Steps Using Derivative Rule for Sum

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

$$\frac{\mathrm{d}}{\mathrm{d}n}(\left(-2\right)^{4}+4\times 2^{3}-2^{3}+n\times 2^{2}-1)$$

Calculate -2 to the power of 4 and get 16.

$$\frac{\mathrm{d}}{\mathrm{d}n}(16+4\times 2^{3}-2^{3}+n\times 2^{2}-1)$$

Calculate 2 to the power of 3 and get 8.

$$\frac{\mathrm{d}}{\mathrm{d}n}(16+4\times 8-2^{3}+n\times 2^{2}-1)$$

Multiply 4 and 8 to get 32.

$$\frac{\mathrm{d}}{\mathrm{d}n}(16+32-2^{3}+n\times 2^{2}-1)$$

Add 16 and 32 to get 48.

$$\frac{\mathrm{d}}{\mathrm{d}n}(48-2^{3}+n\times 2^{2}-1)$$

Calculate 2 to the power of 3 and get 8.

$$\frac{\mathrm{d}}{\mathrm{d}n}(48-8+n\times 2^{2}-1)$$

Subtract 8 from 48 to get 40.

$$\frac{\mathrm{d}}{\mathrm{d}n}(40+n\times 2^{2}-1)$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\mathrm{d}}{\mathrm{d}n}(40+n\times 4-1)$$

Subtract 1 from 40 to get 39.

$$\frac{\mathrm{d}}{\mathrm{d}n}(39+n\times 4)$$

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}.

$$4n^{1-1}$$

Subtract 1 from 1.

$$4n^{0}$$

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

$$4\times 1$$

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

$$4$$