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

1 Answer
Oct 29, 2016

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))