# find `dy/dx` for `y=log[lnx]`

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### 1 Answer

`y=log[lnx]`

To take the derivative of y with respect to x, apply the rule `(log_b u)'= 1/(u lnb)*u'` .

Since the base of logarithm (log) is not written, it indicates that its base is 10.

`y=log_10[lnx]`

So,

`dy/dx=1/(lnx*ln10)*(lnx)'`

Then, apply the rule `(lnu)'=1/u*u'` .

`dy/dx=1/(lnx*ln10)*1/x*x'`

`dy/dx=1/(lnx*ln10)*1/x*1`

`dy/dx=1/(xln10lnx)`

**Hence, `dy/dx=1/(xln10lnx)` .**