What is the derivative of #y= ln(1-x^2)^(1/2)#?

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Answer 1 · 2016-10-29T14:38:56

#y=ln(1-x^2)^(1/2)#

#y^2=ln(1-x^2)# *To be continued...

Differentiate the left hand side of the equation using implicit differentiation. Differentiate the right hand side of the equation using the rule:

If #g(x)=ln(f(x))#, #g'(x)=(f'(x))/f(x)#

*Continued...

#2y*(dy)/(dx)=-(2x)/(1-x^2)#

#(dy)/(dx)=-(2x)/(1-x^2)*1/(2y)#

#(dy)/(dx)=-x/(1-x^2)*1/y#

#(dy)/(dx)=-x/((1+x)(1-x))*1/ln(1-x^2)^(1/2)#

#(dy)/(dx)=-x/(ln(1-x^2)^(1/2)*(1+x)(1-x))#

#(dy)/(dx)=-x/(ln((1+x)(1-x))^(1/2)*(1+x)(1-x))#

#(dy)/(dx)=-x/(sqrt(ln(1+x)+ln(1-x))*(1+x)(1-x))#

To tidy things up a bit, you can multiply the fraction to the right by one, which is the same as #(-1)/(-1)#.

So what you end up with is...

#(dy)/(dx)=x/(sqrt(ln(1+x)+ln(1-x))*(x-1)(x+1))#