differentiate: ln(sqrt((b^2-z^2)/(b^2+z^2)))

Question

differentiate: ln(sqrt((b^2-z^2)/(b^2+z^2)))

Answer 1

I will set this up for you, and you can fill in the math.

 

It may help to do a substitution:

 

Let u(z) = (b² - z²) / (b² + z²)

 

f(z) = ln(√u)

 

Now apply the chain rule.

 

f'(z) = (1 / √u) (1/2) u^-1/2 du/dz = [1 / (2u)] du/dz

 

Substitute u(z) back into this expression, and simplify.  Use the quotient rule to get du/dz.

Answer 2

Assuming that b is a constant, we have:  

 

y = ln(√[(b²-z²)/(b²+z²)])

 

   = ln([(b²-z²)/(b²+z²)]^½

 

   = ½[ln(b²-z²) - ln(b²+z²)]

 

y' = ½[-2z/(b²-z²) - 2z/(b²+z²)]

 

   = -z/(b²-z²) - z/(b²+z²)

 

   = [-z(b²+z²)-z(b²-z²)]/[(b²-z²)(b²+z²)]

 

   = -2zb²/[(b²-z²)(b²+z²)]