# explain the steps used to find `dy/dx` using logarithmic differentiation; `root(9)((x-5)/(x+6))`

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Logarithmic differentiation provides a way to differentiate a function possessing complicated exponents by taking logarithms first, and then by employing the logarithmic and other deriving techniques.

Let `y=root(9)((x-5)/(x+6))=((x-5)/(x+6))^(1/9)`

Taking log of both sides, note the properties of logarithms such that, `log(u/v)=logu-logv` and `log(u)^v=vlogu`

`logy=1/9 log((x-5)/(x+6))=1/9[log(x-5)-log(x+6)]`

Differentiate both sides with respect to x, note that both side requires chain rule of differentiation,

`1/y*dy/dx=1/9[1/(x-5)*1-1/(x+6)*1]`

Multiplying both sides by y such that y is eliminated from the L.H.S,

`rArr dy/dx=1/9*y*[1/(x-5)-1/(x+6)]`

`=1/9*root(9)((x-5)/(x+6))[(x+6-x+5)/((x-5)(x+6))]`

`=11/(9*root(9)((x-5)^8(x+6)^10))`

Therefore, logarithmic differentiation of the given function yields `dy/dx=11/(9*root(9)((x-5)^8(x+6)^10))`

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