Question

Question #17a50

Answer 1

`y' = - frac(1)(2 sqrt(x^(2) - 1))`

Explanation:

We have: `y = ln(sqrt(x - 1) - sqrt(x + 1))`

`Rightarrow y = ln((x - 1)^(frac(1)(2)) - (x + 1)^(frac(1)(2)))`

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

Let `u = (x - 1)^(frac(1)(2)) - (x + 1)^(frac(1)(2))` and `v = ln(u)`:

`Rightarrow y' = u' cdot v'`

`Rightarrow y' = (frac(1)(2) (x - 1)^(- frac(1)(2)) - frac(1)(2) (x + 1)^(- frac(1)(2))) cdot frac(1)(u)`

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

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

`Rightarrow y' = frac(1)(2 u) (frac(sqrt(x + 1))(sqrt(x + 1) sqrt(x - 1)) - frac(sqrt(x - 1))(sqrt(x + 1) sqrt(x - 1)))`

`Rightarrow y' = frac(1)(2 u) (frac(sqrt(x + 1) - sqrt(x - 1))(sqrt(x + 1) sqrt(x - 1)))`

Let's replace `u` with `(x - 1)^(frac(1)(2)) - (x + 1)^(frac(1)(2))`:

`Rightarrow y' = frac(1)(2 ((x - 1)^(frac(1)(2)) - (x + 1)^(frac(1)(2)))) (frac(sqrt(x + 1) - sqrt(x - 1))(sqrt(x + 1) sqrt(x - 1)))`

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

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

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

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

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

`Rightarrow y' = - frac(1)(2 sqrt(x^(2) - 1))` `\ ` (shown.)

Answer 2

Shown in the explanation.

Explanation:

Use the chain rule:

`dy/dx = (d(y(g(x))))/dx = dy/(dg)(dg)/dx" [1]"`

Let `g(x)=sqrt(x-1)-sqrt(x+1)`

`(dg)/dx = 1/(2sqrt(x-1))-1/(2sqrt(x+1))`

Make a common denominator:

`(dg)/dx = 1/(2sqrt(x-1))(sqrt(x+1)/sqrt(x+1))-1/(2sqrt(x+1))(sqrt(x-1)/sqrt(x-1))`

Perform the multiplication:

`(dg)/dx = sqrt(x+1)/(2sqrt(x^2-1))-sqrt(x-1)/(2sqrt(x^2-1))`

Combine over a single denominator:

`(dg)/dx = (sqrt(x+1)-sqrt(x-1))/(2sqrt(x^2-1))`

Multiply by -1:

`(dg)/dx = -(sqrt(x-1)-sqrt(x+1))/(2sqrt(x^2-1))" [2]"`

`y(g) = ln(g)`

`dy/(dg) = 1/g`

Reverse the substitution for g:

`dy/(dg) = 1/(sqrt(x-1)-sqrt(x+1))" [3]"`

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

`dy/dx = (d(y(g(x))))/dx = 1/(sqrt(x-1)-sqrt(x+1))(-(sqrt(x-1)-sqrt(x+1)))/(2sqrt(x^2-1))`

Please observe how the factors cancel:

`dy/dx = (d(y(g(x))))/dx = 1/cancel(sqrt(x-1)-sqrt(x+1))(-cancel(sqrt(x-1)-sqrt(x+1)))/(2sqrt(x^2-1))`

The equation with the factors removed is:

`dy/dx = -1/(2sqrt(x^2-1))`

Q.E.D.