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Differentiate H(z)=ln radical a^2-z^2/radical a^2+z^2

Differentiate 
H(z)=ln radical a^2-z^2/radical a^2+z^2     Note:  Both the denominator and numerator are under radical 
 
 
Differentiate   y=x^lnx
 
Thanks! :))

Comments

What variable do you need to differential with respect to?
All I know is that we need to differentiate and the question is   H(z)= ln radical a^2-z^2/radical a^2+z^2  
 
For the second question, I should also differentiate  y=x^lnx
 
 
In the question it just tells me to "differentiate" and then there is H(z)=  ln radical a^2-z^2/radical a^2+z^2    and
for the second one is   y=x^lnx
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1 Answer

H(z) = ln [√(a2-z2)/√(a2+z2)]
 
       = ln(a2-z2)½  - ln(a2+z2)½
 
       = (½)ln(a2-z2) - (½)ln(a2+z2)
 
H'(z) = -z/(a2-z2) -z/(a2+z2)
 
        = [-z(a2+z2) - z(a2-z2)]/(a4-z4) = -2a2z / (a4-z4)
 
------------------------------------------------------------------------------
y = xlnx
 
Use logarithmic differentiation to find y':
 
lny = ln(xlnx)
 
lny = (lnx)2
 
y'/y = 2(lnx)/x
 
y' = y(2lnx/x) = xlnx(2lnx/x) = 2lnx (xlnx - 1
 

Comments

Thanks for your response.
 
I have a question, how did you get the derivative of (½)ln(a2-z2) - (½)ln(a2+z2)?  I don't understand how you got H'(z) = -z/(a2-z2) -z/(a2+z2)
 
 Thanks again! :))
The derivative of y = ln(u) is u'/u
 
Assuming that a is a constant, the derivative of (1/2)ln(a2+z2) is
 
       (1/2) [ 2z / (a2+z2) ]  = z/(a2+z2)