how would i find the y = version of this expression?

Question

i need to turn x^y = y^x into a y = version so that i can take its derivative. how would i go about this?

Answer 1

I am not sure that an explicit y = f(x) can be found for this problem, but the derivative can still be found for this problem.

 

ln(x^y) = ln(y^x)

ylnx = xlny

 

Taking derivative of both sides with respect to x where y' = dy/dx

y'(lnx) + y(1/x) = lny + x(1/y)y'

(lnx - x/y)y' = lny - y/x

 

y' = (lny - y/x)/(lnx -x/y)

 

Answer 2

Take natural logarithm of both sides.

 

ln(x^y) = ln(y^x)

 

y ln(x) = x ln(y)

 

Now do implicit differentiation, using the chain rule.

 

y' ln(x) + y(1/x) = (1)ln(y) + x(1/y)y'

 

y' ln(x) + y/x = ln(y) + xy'/y

 

y'[ln(x) - x/y] = ln(y) - y/x

 

y'(x,y) = [ln(y) - y/x] / [ln(x) - x/y]

 

Now you need the particular point (x₁,y₁) where you want to evaluate y'.

 

y'(x₁,y₁) = [ln(y₁) - y₁/x₁] / [ln(x₁) - x₁/y₁]

 

I am not sure we can do much more without a pair of coordinates.