Shailesh K. answered 09/28/20
MS in Electrical Engineering with 8+ years of teaching experience
It is work in progress. I do not have answer yet.
I have solved derivative of the function. I don't know the integration yet.
y = x^x^x^x....(r-1) times
ln y = ln x^x^x^x....(r-1) times,
ln y = x^x^x^x…(r-2) times ln x
ln y = y^(1/x) ln x ----------(1) differentiate with respect to x
1/y dy/dx = y^[(1/x)-1] dy/dx (-1/x^2) ln x – y^[(1/x)-1] implicit differentiation
[1/y + y(1/x)-3ln x] dy/dx = – y^[(1/x)-1]
Finally dy/dx = – y^[(1/x) - 1]/ [1/y + y(1/x)-3ln x]
A solution that is very difficult to verify.
I found an answer to integrate the function. I did receive some help from an integration calculator.
For integration let us begin from equation (1)
ln y = y^(1/x) ln x
Let us integrate left side with respect to y
G(y) = ∫ ln y dy = y ln y – y From table of integrals, Standard functions.
Let us integrate right side with respect to x
Let ∫ f(x) dx = F(x) = ∫ y^(1/x) ln x dx integrate by parts ∫ u dv = u v - ∫u’ v
Let u = ln (x) and dv = y^(1/x) differentiate u and integrate dv we get
u’ = 1/x and v = ∫ y^1/x dx = x y^ (1 + 1/x) / (x+1) Substitute these in parts formula
F(x) = ln x · x y^ (1 + 1/x) / (x+1) - ∫ (1/x) [x y^ (1 + 1/x) / (x+1)] dx
= ln x · x y^ (1 + 1/x) / (x+1) - x y^ (2 + 1/x) / [(x+1) (2x + 1)]
= x y^ (1 + 1/x) / (x+1) [ ln x – y / (2x + 1)]
Next, we will simplify equating G(y) = F(x)
y ln y – y = x y^ (1 + 1/x) / (x+1) [ ln x – y / (2x + 1)] Divide by y both sides
ln y -1 = x y^ (1/x) / (x+1) [ ln x – y / (2x + 1)]
ln y = 1 + x y^ (1/x) / (x+1) [ ln x – y / (2x + 1)] Take antilogarithm both sides
y = e^1 + x y^ (1/x) / (x+1) [ ln x – y / (2x + 1)] and finally
y = e e^ x y(1/x) (x+1)-1 [ ln x – y (2x + 1)-1] + C
I will appreciate if any tutor can validate my answers.