**radical**a^2-z^2/

**radical**a^2+z^2 Note: Both the denominator and numerator are under radical

Differentiate

H(z)=ln **radical** a^2-z^2/**radical** a^2+z^2 Note: Both the denominator and numerator are under radical

Differentiate y=x^lnx

Thanks! :))

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H(z) = ln [√(a^{2}-z^{2})/√(a^{2}+z^{2})]

= ln(a^{2}-z^{2})^{½ } - ln(a^{2}+z^{2})^{½}

= (½)ln(a^{2}-z^{2}) - (½)ln(a^{2}+z^{2})

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

= [-z(a^{2}+z^{2}) - z(a^{2}-z^{2})]/(a^{4}-z^{4}) = -2a^{2}z / (a^{4}-z^{4})

------------------------------------------------------------------------------

y = x^{lnx}

Use logarithmic differentiation to find y':

lny = ln(x^{lnx})

lny = (lnx)^{2}

y'/y = 2(lnx)/x

y' = y(2lnx/x) = x^{lnx}(2lnx/x) = 2lnx (x^{lnx - 1})

Thanks for your response.

I have a question, how did you get the derivative of (½)ln(a2-z2) - (½)ln(a2+z2)? I don't understand how you got H'(z) = -z/(a2-z2) -z/(a2+z2)

Thanks again! :))

Thanks again! :))

The derivative of y = ln(u) is u'/u

Assuming that a is a constant, the derivative of (1/2)ln(a^{2}+z^{2}) is

(1/2) [ 2z / (a^{2}+z^{2}) ] = z/(a^{2}+z^{2})

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## Comments

H(z)= ln radical a^2-z^2/radical a^2+z^2 andfor the second one is y=x^lnx