Question

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

Answer 1

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)`