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Grade: 10


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

6 years ago

Answers : (4)

Prashik Raut
21 Points
							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
6 years ago
Nirmal Singh.
askIITians Faculty
44 Points
							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
6 years ago
15 Points
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=
2 years ago
ankit singh
askIITians Faculty
596 Points
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)
4 months ago
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