`H(z) = ln(sqrt((a^2 - z^2)/(a^2 + z^2)))` Differentiate the function.

Textbook Question

Chapter 3, 3.6 - Problem 20 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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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))`






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sbernabel | Student, Graduate | (Level 1) Adjunct Educator

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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)

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