How can I differientiate this fuction? H(z)=ln(√ (b^2-z^2)/(b^2+z^2)

1 Answer
Skewd
Jun 21, 2018

First use the laws of logarithms to rewrite the function as sums and differences of logarithms.
#H(z)=(1/2)ln(b^2-z^2)-(1/2)ln(b^2+z^2)#
#H'(z)=(1/2)(1/(b^2-z^2))(-2z)-(1/2)(1/(b^2+z^2))(2z)#

Explanation:

Remember to use the chain rule.
#(d/dx)lnu=(1/u)((du)/dx)#

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