If # y = (tan^(-1)x)^2 # then show that #(1+x^2)^2y'' + 2x (1+x^2) y' = 2#?

1 Answer
May 20, 2017

To proceed we will need some standard Calculus results:

# d/dx tan^(-1)x = 1/(1+x^2) #

Now we have:

# y = (tan^(-1)x)^2 #

If we apply the chain rule then we get:

# y' = 2*(tan^(-1)x)*1/(1+x^2) #
# \ \ \ \ = (2tan^(-1)x)/(1+x^2) #

And differentiating again and applying the quotient rule, along with the chain rule, we get:

# y'' = { (1+x^2)(d/dx 2tan^(-1)x) - (2tan^(-1)x)(d/dx (1+x^2)) } / (1+x^2)^2 #

# \ \ \ \ = { (1+x^2)(2/(1+x^2)) - (2tan^(-1)x)(2x) } / (1+x^2)^2 #

# \ \ \ \ = { 2 - 2x * (2tan^(-1)x) } / (1+x^2)^2 #

# :. (1+x^2)^2y'' = 2 - 2x * (2tan^(-1)x) #

# :. (1+x^2)^2y'' = 2 - 2x (1+x^2)* (2tan^(-1)x)/(1+x^2) #

# :. (1+x^2)^2y'' = 2 - 2x (1+x^2) y' #

# :. (1+x^2)^2y'' + 2x (1+x^2) y' = 2 # QED