Solve 2(2)^6+2(2)^4-x^2+1

$$2 ( 2 ) ^ { 6 } + 2 ( 2 ) ^ { 4 } - x ^ { 2 } + 1$$

Evaluate

$$161-x^{2}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 1 and 6 to get 7.

$$2^{7}+2\times 2^{4}-x^{2}+1$$

To multiply powers of the same base, add their exponents. Add 1 and 4 to get 5.

$$2^{7}+2^{5}-x^{2}+1$$

Calculate 2 to the power of 7 and get 128.

$$128+2^{5}-x^{2}+1$$

Calculate 2 to the power of 5 and get 32.

$$128+32-x^{2}+1$$

Add 128 and 32 to get 160.

$$160-x^{2}+1$$

Add 160 and 1 to get 161.

$$161-x^{2}$$

Differentiate w.r.t. x

$$-2x$$

Steps Using Derivative Rule for Sum

To multiply powers of the same base, add their exponents. Add 1 and 6 to get 7.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2^{7}+2\times 2^{4}-x^{2}+1)$$

To multiply powers of the same base, add their exponents. Add 1 and 4 to get 5.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2^{7}+2^{5}-x^{2}+1)$$

Calculate 2 to the power of 7 and get 128.

$$\frac{\mathrm{d}}{\mathrm{d}x}(128+2^{5}-x^{2}+1)$$

Calculate 2 to the power of 5 and get 32.

$$\frac{\mathrm{d}}{\mathrm{d}x}(128+32-x^{2}+1)$$

Add 128 and 32 to get 160.

$$\frac{\mathrm{d}}{\mathrm{d}x}(160-x^{2}+1)$$

Add 160 and 1 to get 161.

$$\frac{\mathrm{d}}{\mathrm{d}x}(161-x^{2})$$

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}.

$$2\left(-1\right)x^{2-1}$$

Multiply 2 times -1.

$$-2x^{2-1}$$

Subtract 1 from 2.

$$-2x^{1}$$

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

$$-2x$$