What is the derivative of #ln((x^2-4)/(2x+5))#?

1 Answer
Euan S.
Jul 30, 2016

#(dy)/(dx) = (2x^2+10x+8)/((x^2-4)(2x+5))#

Explanation:

We have a function y(u(x)) so we need to use the chain rule.

#(dy)/(dx) = (dy)/(du)(du)/(dx)#

Set #u = (x^2-4)/(2x+5)# and using the quotient rule obtain:

#(du)/(dx) = (2x(2x+5) - 2(x^2-4))/(2x+5)^2 = (2x^2+10x+8)/(2x+5)^2#

#y = ln(u) implies (dy)/(du) = 1/u = (2x+5)/(x^2-4)#

#therefore (dy)/(dx) = (2x^2+10x+8)/((x^2-4)(2x+5))#

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