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)-3}ln x] dy/dx = – y^[(1/x)-1]

Finally dy/dx = – y^[(1/x) - 1]/ [1/y + y^{(1/x)-3}ln 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.