# Question 17a50

May 19, 2017

$y ' = - \frac{1}{2 \sqrt{{x}^{2} - 1}}$

#### Explanation:

We have: $y = \ln \left(\sqrt{x - 1} - \sqrt{x + 1}\right)$

$R i g h t a r r o w y = \ln \left({\left(x - 1\right)}^{\frac{1}{2}} - {\left(x + 1\right)}^{\frac{1}{2}}\right)$

This function can be differentiated using the "chain rule".

Let $u = {\left(x - 1\right)}^{\frac{1}{2}} - {\left(x + 1\right)}^{\frac{1}{2}}$ and $v = \ln \left(u\right)$:

$R i g h t a r r o w y ' = u ' \cdot v '$

$R i g h t a r r o w y ' = \left(\frac{1}{2} {\left(x - 1\right)}^{- \frac{1}{2}} - \frac{1}{2} {\left(x + 1\right)}^{- \frac{1}{2}}\right) \cdot \frac{1}{u}$

$R i g h t a r r o w y ' = \frac{1}{u} \cdot \left(\frac{1}{2 \sqrt{x - 1}} - \frac{1}{2 \sqrt{x + 1}}\right)$

$R i g h t a r r o w y ' = \frac{1}{u} \cdot \left(\frac{1}{2} \cdot \left(\frac{1}{\sqrt{x - 1}} - \frac{1}{\sqrt{x + 1}}\right)\right)$

$R i g h t a r r o w y ' = \frac{1}{2 u} \left(\frac{\sqrt{x + 1}}{\sqrt{x + 1} \sqrt{x - 1}} - \frac{\sqrt{x - 1}}{\sqrt{x + 1} \sqrt{x - 1}}\right)$

$R i g h t a r r o w y ' = \frac{1}{2 u} \left(\frac{\sqrt{x + 1} - \sqrt{x - 1}}{\sqrt{x + 1} \sqrt{x - 1}}\right)$

Let's replace $u$ with ${\left(x - 1\right)}^{\frac{1}{2}} - {\left(x + 1\right)}^{\frac{1}{2}}$:

$R i g h t a r r o w y ' = \frac{1}{2 \left({\left(x - 1\right)}^{\frac{1}{2}} - {\left(x + 1\right)}^{\frac{1}{2}}\right)} \left(\frac{\sqrt{x + 1} - \sqrt{x - 1}}{\sqrt{x + 1} \sqrt{x - 1}}\right)$

Rightarrow y' = frac(sqrt(x + 1) - sqrt(x - 1))(2 cdot (sqrt(x - 1) - sqrt(x + 1)) cdot sqrt(x + 1) sqrt(x - 1)))#

$R i g h t a r r o w y ' = \frac{\left(\sqrt{x + 1} - \sqrt{x - 1}\right)}{2 \cdot - 1 \cdot \left(\sqrt{x + 1} - \sqrt{x - 1}\right) \cdot \sqrt{x + 1} \sqrt{x - 1}}$

$R i g h t a r r o w y ' = - \frac{1}{2 \cdot \sqrt{x + 1} \sqrt{x - 1}}$

$R i g h t a r r o w y ' = - \frac{1}{2 \sqrt{\left(x + 1\right) \left(x - 1\right)}}$

The argument of the radical can be factorised as a difference of two squares:

$R i g h t a r r o w y ' = - \frac{1}{2 \sqrt{{x}^{2} - 1}}$ $\setminus$ (shown.)

May 19, 2017

Shown in the explanation.

#### Explanation:

Use the chain rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d \left(y \left(g \left(x\right)\right)\right)}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dg}} \frac{\mathrm{dg}}{\mathrm{dx}} \text{ [1]}$

Let $g \left(x\right) = \sqrt{x - 1} - \sqrt{x + 1}$

$\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x - 1}} - \frac{1}{2 \sqrt{x + 1}}$

Make a common denominator:

$\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{1}{2 \sqrt{x - 1}} \left(\frac{\sqrt{x + 1}}{\sqrt{x + 1}}\right) - \frac{1}{2 \sqrt{x + 1}} \left(\frac{\sqrt{x - 1}}{\sqrt{x - 1}}\right)$

Perform the multiplication:

$\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{\sqrt{x + 1}}{2 \sqrt{{x}^{2} - 1}} - \frac{\sqrt{x - 1}}{2 \sqrt{{x}^{2} - 1}}$

Combine over a single denominator:

$\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{\sqrt{x + 1} - \sqrt{x - 1}}{2 \sqrt{{x}^{2} - 1}}$

Multiply by -1:

$\frac{\mathrm{dg}}{\mathrm{dx}} = - \frac{\sqrt{x - 1} - \sqrt{x + 1}}{2 \sqrt{{x}^{2} - 1}} \text{ [2]}$

$y \left(g\right) = \ln \left(g\right)$

$\frac{\mathrm{dy}}{\mathrm{dg}} = \frac{1}{g}$

Reverse the substitution for g:

$\frac{\mathrm{dy}}{\mathrm{dg}} = \frac{1}{\sqrt{x - 1} - \sqrt{x + 1}} \text{ [3]}$

Substitute equations [2] and [3] into equation [1]:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d \left(y \left(g \left(x\right)\right)\right)}{\mathrm{dx}} = \frac{1}{\sqrt{x - 1} - \sqrt{x + 1}} \frac{- \left(\sqrt{x - 1} - \sqrt{x + 1}\right)}{2 \sqrt{{x}^{2} - 1}}$

Please observe how the factors cancel:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d \left(y \left(g \left(x\right)\right)\right)}{\mathrm{dx}} = \frac{1}{\cancel{\sqrt{x - 1} - \sqrt{x + 1}}} \frac{- \cancel{\sqrt{x - 1} - \sqrt{x + 1}}}{2 \sqrt{{x}^{2} - 1}}$

The equation with the factors removed is:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \sqrt{{x}^{2} - 1}}$

Q.E.D.