How can I differentiate the function?

H(z)=#ln(sqrt(b^2-z^2)/(b^2+z^2))#

1 Answer
Mar 2, 2018

#H'(z)=-z/(b^2-z^2)-(2z)/(b^2+z^2)#

Explanation:

We want to find the derivative of

#H(z)=ln((sqrt(b^2-z^2))/(b^2+z^2))#

Simplify using the properties of logarithms

#H(z)=ln(sqrt(b^2-z^2))-ln(b^2+z^2)#

#=1/2ln(b^2-z^2)-ln(b^2+z^2)#

Differentiate using the chain rule

#H'(z)=1/2(-2z)1/(b^2-z^2)-(2z)/(b^2+z^2)#

#=-z/(b^2-z^2)-(2z)/(b^2+z^2)#