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
