Grade 10Differential Calculus

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

Arpit Kumar

12 Years agoGrade 10

4 Answers

Prashik Raut

Approved Tutor Answer12 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.

Approved Tutor Answer12 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

Pallab

7 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^...is 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

I/y*d(y)/dx=y/x+logx*d(y)/dx

Bring d(y)/dx on one side

Then take d(y)/dx common

Finally take the remaining part on other side you will get=

d(y)/dx=y^2/x(1-ylogx).

ankit singh

5 Years 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)