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

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! :))


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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Mark M. | Math Tutor--High School/College levelsMath Tutor--High School/College levels
4.9 4.9 (821 lesson ratings) (821)
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


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)