What's the error in this solution?

Question

This is my textbook problem. I solved it with three methods:
A) By rationalising the denominator {WRONG}
B) By writing it as (denominator)^-1 and using the power rule {RIGHT}
C) By using the quotient rule {RIGHT}

I also verified it using online derivative calculators. And it turns out A is incorrect.

Can someone explain what the error is, in the work shown below, using method A?

Answer 1

There is no error when you rationalized the denominator before taking the derivative. Instead, it's just that it's expressed in a different form. For example, with your second method, i.e.,

B) By writing it as (denominator)^-1 and using the power rule {RIGHT}

This can be written as

$$f(x) = \left(\sqrt{x + 1} - \sqrt{x - 1}\right)^{-1} \tag{1}\label{eq1A}$$

so you then have

$$\begin{equation}\begin{aligned} f'(x) & = -\left(\sqrt{x + 1} - \sqrt{x - 1}\right)^{-2}\left(\frac{1}{2\sqrt{x + 1}} - \frac{1}{2\sqrt{x - 1}}\right) \\ & = \left(\frac{1}{2}\right)\left(\frac{1}{\left(\sqrt{x + 1} - \sqrt{x - 1}\right)^2}\right)\left(\frac{1}{\sqrt{x - 1}} - \frac{1}{\sqrt{x + 1}}\right) \\ & = \left(\frac{1}{2}\right)\left(\frac{1}{\left(\sqrt{x + 1} - \sqrt{x - 1}\right)^2}\right)\left(\frac{\sqrt{x+1} - \sqrt{x - 1}}{\sqrt{x - 1}\sqrt{x + 1}}\right) \\ & = \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{x + 1} - \sqrt{x - 1}}\right)\left(\frac{1}{\sqrt{x - 1}\sqrt{x + 1}}\right) \\ & = \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{x - 1}\sqrt{x + 1}}\right)\left(\frac{1}{\sqrt{x + 1} - \sqrt{x - 1}}\right)\left(\frac{\sqrt{x + 1} + \sqrt{x - 1}}{\sqrt{x + 1} + \sqrt{x - 1}}\right) \\ & = \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{x - 1}\sqrt{x + 1}}\right)\left(\frac{\sqrt{x + 1} + \sqrt{x - 1}}{(x + 1) - (x - 1)}\right) \\ & = \left(\frac{1}{4}\right)\left(\frac{\sqrt{x + 1} + \sqrt{x - 1}}{\sqrt{x - 1}\sqrt{x + 1}}\right) \\ & = \left(\frac{1}{4}\right)\left(\frac{\sqrt{x + 1}}{\sqrt{x - 1}\sqrt{x + 1}} + \frac{\sqrt{x - 1}}{\sqrt{x - 1}\sqrt{x + 1}}\right) \\ & = \left(\frac{1}{4}\right)\left(\frac{1}{\sqrt{x-1}} + \frac{1}{\sqrt{x+1}}\right) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

As you can see, it's the same result as what you got using your method A. Your your third method would get the same derivative as the first line in \eqref{eq2A}, so it would also match the result of your method A.