Solve (sqrt{1+x}-sqrt{1-x})*(sqrt{1+x}+sqrt{1-x})

$$( \sqrt{ 1+x } - \sqrt{ 1-x } ) \times ( \sqrt{ 1+x } + \sqrt{ 1-x } )$$

Evaluate

$$2x$$

Solution Steps

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate \sqrt{1+x} to the power of 2 and get 1+x.

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

Calculate \sqrt{1-x} to the power of 2 and get 1-x.

$$1+x-\left(1-x\right)$$

To find the opposite of 1-x, find the opposite of each term.

$$1+x-1-\left(-x\right)$$

The opposite of -x is x.

$$1+x-1+x$$

Subtract 1 from 1 to get 0.

$$x+x$$

Combine x and x to get 2x.

$$2x$$

Differentiate w.r.t. x

$$2$$

Steps Using Definition of a Derivative

Consider \left(\sqrt{1+x}-\sqrt{1-x}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(\sqrt{1+x}\right)^{2}-\left(\sqrt{1-x}\right)^{2})$$

Calculate \sqrt{1+x} to the power of 2 and get 1+x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(1+x-\left(\sqrt{1-x}\right)^{2})$$

Calculate \sqrt{1-x} to the power of 2 and get 1-x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(1+x-\left(1-x\right))$$

To find the opposite of 1-x, find the opposite of each term.

$$\frac{\mathrm{d}}{\mathrm{d}x}(1+x-1-\left(-x\right))$$

The opposite of -x is x.

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

Subtract 1 from 1 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x+x)$$

Combine x and x to get 2x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x)$$

The derivative of ax^{n} is nax^{n-1}.

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

Subtract 1 from 1.

$$2x^{0}$$

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

$$2\times 1$$

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

$$2$$