Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx((log(((1 + x)/(-1 + x))^log(3)))/log(3))
Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):
d/dx(x) y'(x) = d/dx((log(((1 + x)/(-1 + x))^log(3)))/log(3))
The derivative of x is 1:
1 y'(x) = d/dx((log(((1 + x)/(-1 + x))^log(3)))/log(3))
Factor out constants:
y'(x) = (d/dx(log(((1 + x)/(-1 + x))^log(3))))/(log(3))
Simplify log(((x + 1)/(x - 1))^log(3)) using the identity log(a^b) = b log(a):
y'(x) = (d/dx(log(3) log((1 + x)/(-1 + x))))/log(3)
Factor out constants:
y'(x) = (log(3) d/dx(log((1 + x)/(-1 + x))))/log(3)
Simplify the expression:
y'(x) = d/dx(log((1 + x)/(-1 + x)))
Simplify log((x + 1)/(x - 1)) using the identity log(a/b) = log(a) - log(b):
y'(x) = d/dx(log(1 + x) - log(-1 + x))
Differentiate the sum term by term and factor out constants:
y'(x) = d/dx(log(1 + x)) - d/dx(log(-1 + x))
Using the chain rule, d/dx(log(x - 1)) = ( dlog(u))/( du) ( du)/( dx), where u = x - 1 and d/( du)(log(u)) = 1/u:
y'(x) = d/dx(log(1 + x)) - (d/dx(-1 + x))/(x - 1)
Differentiate the sum term by term:
y'(x) = d/dx(log(1 + x)) - (d/dx(-1) + d/dx(x))/(x - 1)
The derivative of -1 is zero:
y'(x) = d/dx(log(1 + x)) - (d/dx(x) + 0)/(-1 + x)
Simplify the expression:
y'(x) = -(d/dx(x))/(-1 + x) + d/dx(log(1 + x))
The derivative of x is 1:
y'(x) = d/dx(log(1 + x)) - 1/(x - 1)
Using the chain rule, d/dx(log(x + 1)) = ( dlog(u))/( du) ( du)/( dx), where u = x + 1 and d/( du)(log(u)) = 1/u:
y'(x) = -1/(-1 + x) + (d/dx(1 + x))/(x + 1)
Differentiate the sum term by term:
y'(x) = -1/(-1 + x) + (d/dx(1) + d/dx(x))/(x + 1)
The derivative of 1 is zero:
y'(x) = -1/(-1 + x) + (d/dx(x) + 0)/(1 + x)
Simplify the expression:
y'(x) = -1/(-1 + x) + (d/dx(x))/(1 + x)
The derivative of x is 1:
y'(x) = -1/(-1 + x) + 1/(x + 1) - Courtesy of Mathematica 11 Home Edition