Question

Derivative of x^(lnx)^ln(lnx)?

Answer 1

`x^((lnx)^(ln(lnx))-1)(lnx)^ln(lnx)(2ln(lnx)+1)`

Explanation:

`y=x^((lnx)^(ln(lnx)))`

Take the natural log of both sides:

`lny=ln(x^((lnx)^(ln(lnx))))=(lnx)^ln(lnx)(lnx)=(lnx)^(ln(lnx)+1)`

Take the natural log once more:

`ln(lny)=ln((lnx)^(ln(lnx)+1))=(ln(lnx)+1)(ln(lnx))`

Now take the derivative of both sides. Use the chain rule on the left and product and chain rules on the right.

`1/lny(d/dxlny)=(d/dx(ln(lnx)+1))ln(lnx)+(ln(lnx)+1)(d/dxln(lnx))`

`1/lny(1/y)dy/dx=(1/lnxd/dxlnx)ln(lnx)+(ln(lnx)+1)(1/lnxd/dxlnx)`

`1/(ylny)dy/dx=ln(lnx)/(xlnx)+(ln(lnx)+1)/(xlnx)`

Solving for the derivative:

`dy/dx=(ylny(2ln(lnx)+1))/(xlnx)`

Substituting in `y` and `lny` as defined and previously derived:

`dy/dx=(x^((lnx)^(ln(lnx)))(lnx)^(ln(lnx)+1)(2ln(lnx)+1))/(xlnx)`

`dy/dx=x^((lnx)^(ln(lnx))-1)(lnx)^ln(lnx)(2ln(lnx)+1)`

Answer 2

`(dy)/(dx)=(y)/(x)xx(lnx)^(ln(lnx)){2ln(lnx)+1}`

Explanation:

Let,

`y=x^((lnx)^(ln(lnx))`

For simplicity we take,

`N=(lnx)^(ln(lnx))`

So, `y=x^N`

Taking log ,we get

`lny=lnx^N=N*lnx` , where , `N=(lnx)^(ln(lnx))`

`=>lny=(lnx)^(ln(lnx))*lnx...to(A)`

Again taking log,we get

`ln(lny)=ln((lnx)^(ln(lnx))*lnx)`

`=>ln(lny)=ln((lnx)^(ln(lnx)))+ln(lnx)`

`=>ln(lny)=ln(lnx)(ln(lnx)+ln(lnx)`

`=>ln(lny)=ln(lnx)[ln(lnx)+1]`

Diff. w. r. t. `x` ,`"using "color(blue)"product and Chain Rule :"`

`1/(lny)*1/y(dy)/(dx)`=`ln(lnx)[1/lnx*1/x]+[ln(lnx)+1]xx1/lnx1/x`

`=>1/(lny)*1/y(dy)/(dx)`=`[1/lnx*1/x]{ln(lnx)+ln(lnx)+1}`

`=>1/(ylny)(dy)/(dx)`=`[1/(xlnx)]{2ln(lnx)+1}`

`=>(dy)/(dx)=(ylny)/(xlnx){2ln(lnx)+1}`

OR. from `(A)` , we can put ,

`color(blue)(lny=(lnx)^(ln(lnx))*lnx)`

So,

`(dy)/(dx)=(y)/(xlnx)xxcolor(blue)((lnx)^(ln(lnx)) *lnx) {2ln(lnx)+1}`

`=>(dy)/(dx)=(y)/(x)xx(lnx)^(ln(lnx)){2ln(lnx)+1}`