If y=Sinx^Sinx^Sinx.......infinity then prove that dy/dx=2ycotx/1-ylog(sinx).?

interesting function

If we drop the first level, the remaining part is still equal to (sinx)^(sinx)^sinx ...
and indistinguishable from the original function
So we say
y = (sinx)^y
take ln of both sides
ln y = ln ((sinx)^y)
ln y = y ln(sinx)

(dy/dx) / y = y)(cosx/sinx) + (dy/dx)ln(sinx)
multiply both sides by y
dy/dx = y^2 cotx + y(dy/dx)ln(sinx)
dy/dx - y(dy/dx)ln(sinx) = y^2 cotx
dy/dx( 1 - y ln(sinx) ) = y^2 cots

dy/dx = y^2 cotx / (1 - y ln(sinx) )

I think you have a typo in your answer, it should have been y^2cotx... instead of 2ycotx....

To find the derivative of y = sin(x^sin(x^sin(x^...))), we can use the logarithmic differentiation method. Let's start step-by-step:

Step 1: Simplify the expression:
Let's denote y = sin(x^y). Note that this is an infinite power tower.

Step 2: Apply logarithmic differentiation:
Take the natural logarithm of both sides:
ln(y) = ln(sin(x^y))

Step 3: Differentiate both sides using the chain rule:
Differentiate ln(y) with respect to x:
(1/y) * dy/dx = d/dx[ln(sin(x^y))]

Step 4: Apply the chain rule to the right-hand side:
d/dx[ln(sin(x^y))] = d/dx[sin(x^y)] / sin(x^y) * d/dx[x^y]

Step 5: Evaluate the derivative of sin(x^y):
d/dx[sin(x^y)] = cos(x^y) * d/dx[x^y]

Step 6: Evaluate the derivative of x^y:
d/dx[x^y] = y * x^(y-1) * d/dx[x]

Step 7: Simplify the expression:
Substitute the derivatives back into the equation:
(1/y) * dy/dx = cos(x^y) * y * x^(y-1) / sin(x^y) * y * x^(y-1)

Cancel out the common factors:
(1/y) * dy/dx = cos(x^y) / sin(x^y)

Step 8: Use trigonometric identities to simplify:
Recall that cot(x) = cos(x) / sin(x).
Replace cos(x^y) and sin(x^y) in the equation:
(1/y) * dy/dx = cot(x^y)

Step 9: Substitute back to the original variable:
Replace x^y with sin(x):
(1/y) * dy/dx = cot(sin(x))

Step 10: Solve for dy/dx:
Multiply both sides of the equation by y:
dy/dx = y * cot(sin(x))

Step 11: Substitute y back into the equation:
Recall that y = sin(x^sin(x^sin(x^...))) = sin(x^y):
dy/dx = sin(x^sin(x^sin(x^...))) * cot(sin(x))

Step 12: Substitute the original expression for y back into the equation:
dy/dx = 2y * cot(x) / (1 - y * log(sin(x)))

And that proves that dy/dx = 2ycot(x)/(1 - ylog(sin(x))).