Question #17a50

2 Answers
May 19, 2017

#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.)

May 19, 2017

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.