`H(z)=lnsqrt((a^2-z^2)/(a^2+z^2))`

Differentiating both sides with respect to z, we get

`H'(z)=(1/sqrt((a^2-z^2)/(a^2+z^2))) d/dzsqrt((a^2-z^2)/(a^2+z^2))`

`H'(z)=sqrt((a^2+z^2)/(a^2-z^2)) ((sqrt(a^2+z^2)d/dzsqrt(a^2-z^2)-sqrt(a^2-z^2)d/dzsqrt(a^2+z^2))/(a^2+z^2))`

`H'(z)=sqrt((a^2+z^2)/(a^2-z^2)) ((sqrt(a^2+z^2)(1/2)(a^2-z^2)^(-1/2)(-2z)-sqrt(a^2-z^2)(1/2)(a^2+z^2)^(-1/2)(2z))/(a^2+z^2))`

`H'(z)=sqrt((a^2+z^2)/(a^2-z^2))((-zsqrt((a^2+z^2)/(a^2-z^2))-zsqrt((a^2-z^2)/(a^2+z^2)))/(a^2+z^2))`

`H'(z)=(((-z(a^2+z^2))/(a^2-z^2))-z)/(a^2+z^2)`

`H'(z)=(-z(a^2+z^2)-z(a^2-z^2))/((a^2-z^2)(a^2+z^2))`

`H'(z)=(-z(a^2+z^2+a^2-z^2))/(a^4-z^4)`

`H'(z)=(-2a^2z)/(a^4-z^4)`

We have H(z)=ln(sqrt((a^2+z^2)/(a^2-z^2))). We will do the differentiation in steps, using the chain rule. Suppose you have H(z)=f(g(u(z))). Then H'(z)=f'(g(u(z)))*g'(u(z))*u'(z).

In our case f is ln(), which has derivative f'= 1/(). the () is whatever is inside the parenthesis. Then g is sqrt(), and g'=1/2*()^(-1/2).

Finally we have u(z)=(a^2+z^2)/(a^2-z^2). To find the derivative we use the formula (n/d)'=(n'*d-d'*n)/d^2 (this is called the quotient rule), where n is the numerator, and d the denominator. So that, u'(z) = 4*a^2*z/(a^2-z^2)^2.

Now combine the steps to get:

H'(z) = 1/sqrt((a^2+z^2)/(a^2-z^2)) * 1/2*((a^2+z^2)/(a^2-z^2))^(-1/2) * 4*a^2*z/(a^2-z^2)^2

After simplifying,

H'(z) = 2*a^2*z/(a^4-z^4)

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now