Find dy/dx of y=(tan x)^cot x + (cot x)^tan x.

Let y1=(tanx)^cotx

Take log on both sides,

logy1=cotx.logtanx

diff.w.r.to x , get

(1/y1)dy1/dx=cotx.(1/tanx).sec^2x+logtanx.(-cosec^2x)

dy1/dx=(tanx)^cotx .cosec^2x[1- logtanx]

let y2=(cotx)^tanx

taking log on both sides, we get

logy2=tanxlogcotx

diff.w.r.to x , get

1/y2.dy2/dx=tanx.(1/cotx)(-cosec^2x)+logcotx.sec^2x

= (cotx)^tanx (sec^2x)[-1+logcotx]

Hence dy/dx=(tanx)^cotx .cosec^2x[1- logtanx]+ (cotx)^tanx (sec^2x)[-1+logcotx]

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