If #y = tanh^(-1)((2x)/(1+x^2))# then show that the second derivative is #(4x)/(1 -x^2)^2#?
2 Answers
We have:
# y = tanh^(-1)((2x)/(1+x^2)) #
From which we have:
# tanh y = (2x)/(1+x^2) #
Differentiating Implicitly, and applying the chain rule, we get:
# sech^2ydy/dx = ( (1+x^2)(2) -(2x)(2x) ) / (1+x^2)^2 #
# :. (1-tanh^2y) \ dy/dx = ( 2+2x^2 -4x^2 ) / (1+x^2)^2 #
# :. (1-((2x)/(1+x^2))^2) \ dy/dx = ( 2 -2x^2 ) / (1+x^2)^2 #
# :. ( ((1+x^2)^2-4x^2)/(1+x^2)^2) \ dy/dx = ( 2 -2x^2 ) / (1+x^2)^2 #
# :. ( 1+2x^2+x^4-4x^2) \ dy/dx = 2 -2x^2 #
# :. ( 1-2x^2+x^4) \ dy/dx = 2 -2x^2 #
# :. ( 1-x^2)^2 \ dy/dx = 2(1 -x^2) #
# :. dy/dx = 2/(1 -x^2) #
And differentiating again, and applying the chain rule, we gte
# (d^2y)/(dx^2) = 2(-1)(1 -x^2)^(-2)(-2x) #
# \ \ \ \ \ \ \ = (4x)/(1 -x^2)^2 \ \ # QED
We have:
# y = tanh^(-1)((2x)/(1+x^2)) #
Using the standard result:
# d/dx tanhx = 1/(1-x^2) #
In conjunction with the chain rule and quotient rule, then we get:
# dy/dx = 1/(1-((2x)/(1+x^2))^2) d/dx ((2x)/(1+x^2)) #
# \ \ \ \ \ \ = 1/( ( (1+x)^2-4x^2)/((1+x^2)^2)) { ( (1+x^2)(2)-(2x)(2x) ) / (1+x^2)^2} #
# \ \ \ \ \ \ = ((1+x^2)^2)/( ( (1+x^2)^2-4x^2)) { ( (1+x^2)(2)-(2x)(2x) ) / (1+x^2)^2} #
# \ \ \ \ \ \ = 2 \ ( 1+x^2-2x^2 ) / ( ( 1+2x^2+x^4-4x^2)) #
# \ \ \ \ \ \ = 2 \ ( 1-x^2 ) / ( 1-2x^2+x^4) #
# \ \ \ \ \ \ = 2 \ ( 1-x^2 ) / ( (1-x^2)^2) #
# \ \ \ \ \ \ = 2 / ( 1-x^2) #
And differentiating again, and applying the chain rule, we gte
# (d^2y)/(dx^2) = 2(-1)(1 -x^2)^(-2)(-2x) #
# \ \ \ \ \ \ \ = (4x)/(1 -x^2)^2 \ \ # QED
