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

```
6 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
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6 years ago
```							y = x^x^x^x^-------infinitewe can write it-----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)Thanks & Regards,Nirmal SinghAskiitians Faculty
```
6 years ago
```							Let us consider x^x^xupto infinity to be yThen y=x^x^x^x...It gives y=x^y.             (Since x^x^x^...is y)Then we can proceedTake log on both sidesWe getLog y=logx^yLogy=y logxNow differentiated(logy)/dx=d(y logx)/dxNow on RHS apply product rule.We getI/y*d(y)/dx=y/x+logx*d(y)/dxBring d(y)/dx on one side Then take d(y)/dx commonFinally take the remaining part on other side you will get=d(y)/dx=y^2/x(1-ylogx).
```
2 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)
```
4 months ago
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