# y=(ln x)^(cos9x)

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### 1 Answer

You need to take logarithms both sides such that:

`ln y = ln((ln x)^(cos9x))`

Using the logarithmic identities yields:

`ln y = cos9x*ln(ln x)`

You need to differentiate both sides such that:

`(1/y)*y' = (cos9x)'*ln(ln x)+ cos9x*(ln(ln x))'`

`(1/y)*y' = (-9sin9x)*ln(ln x) + cos9x*(1/(ln x))*(1/x)`

`y' = y*((-9sin9x)*ln(ln x) + cos9x*(1/(ln x))*(1/x)) `

Substituting `((ln x)^(cos9x))` for y yields:

`y' = ((ln x)^(cos9x))*((-9sin9x)*ln(ln x) +(cos9x)/(xln x))`

**Hence, differentiating the function with respect to x yields `y' = ((ln x)^(cos9x))*((-9sin9x)*ln(ln x) +(cos9x)/(xln x)).` **