How do you find the second derivative of #y= ln(1-x^2)^(1/2) #?

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Answer 1 · 2016-10-30T00:54:52

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

# y = ln(1-x^2)^(1/2) # can be rewritten as # y^2=ln(1-x^2) #

Differentiating wrt #x# gives:
# d/dx(y^2) = d/dx( ln(1-x^2) ) #

# :. dy/dxd/dy(y^2) = d/dx{ ln(1-x^2) } #

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

# :. ydy/dx = (x)/(x^2-1) #

# :. ln(1-x^2)^(1/2)(dy/dx) = (x)/(x^2-1) #

# :. dy/dx = (x)/((x^2-1)sqrtln(1-x^2)) #