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y=x^x^x^x.... Till infinite Find dy/dx Please give the solution

y=x^x^x^x.... Till infinite
Find dy/dx
Please give the solution


4 Answers

Prashik Raut
21 Points
7 years ago
Write in the form y=x^y since x^x^x^x.. is till infinite, we can consider it to be y, then take log on both sides and differentiate
Nirmal Singh.
askIITians Faculty 44 Points
7 years ago
y = x^x^x^x^-------infinite
we can write it-----
y = x^y
taking log both side
log y = log (x^y)
log y = y log x
taking differentiation both side with respect to x
d/dx (log y) = d/dx (y log x)
1/y*dy/dx = y* d/dx(logx)+logx*dy/dx
1/y*dy/dx = y / x+logx*dy/dx
dy/dx (1/y - logx) = y/x
dy/dx = y^2 / x(1 - y lox)
Thanks & Regards,
Nirmal Singh
Askiitians Faculty
15 Points
3 years ago
Let us consider x^x^xupto infinity to be y
Then y=x^x^x^x...
It gives y=x^y.             (Since x^x^x^ y)
Then we can proceed
Take log on both sides
We get
Log y=logx^y
Logy=y logx
Now differentiate
d(logy)/dx=d(y logx)/dx
Now on RHS apply product rule.
We get
Bring d(y)/dx on one side 
Then take d(y)/dx common
Finally take the remaining part on other side you will get=
ankit singh
askIITians Faculty 614 Points
one year ago
y = x^ytaking log both sidelog y = log (x^y)log y = y log xtaking differentiation both side with respect to xd/dx (log y) = d/dx (y log x)1/y*dy/dx = y* d/dx(logx)+logx*dy/dx1/y*dy/dx = y / x+logx*dy/dxdy/dx (1/y - logx) = y/xdy/dx = y^2 / x(1 - y lox)

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