Question

How do you find the derivative of `(ln(x))^x`?

Answer 1

`dy/dx=(lnx)^x{1/lnx+ln(lnx)}.`

Explanation:

`"Let "y=(lnx)^x.`

Taking Natural Log. of both sides, & using the Rule of Log. Fun, we get,

`lny=xln(lnx)`

Now, diff.ing both sides w.r.t. `x`, we get,

`d/dx(lny)=d/dx{xln(lnx)}`

`=xd/dx{ln(lnx)}+{ln(lnx)}d/dx{x}[because," the Product Rule]"`

Now, using the Chain Rule :

`(1) : d/dx(lny)=d/dy(lny)(d/dx(y))=1/ydy/dx;`

`(2) : d/dx{ln(lnx)}=1/lnxd/dx(lnx)=(1/lnx)(1/x)=1/(xlnx);`

Therefore, taking `(1) and (2)` into account, we finally have,

`(1/y)(dy/dx)=x{1/(xlnx)}+{ln(lnx)}(1)=1/lnx+ln(lnx)`

`dy/dx=y{1/lnx+ln(lnx)},` and since, `y=(lnx)^x,` we have,

`dy/dx=(lnx)^x{1/lnx+ln(lnx)}.`

Enjoy Maths., and, Spread its Joy!