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Grade: 12
        
find dy/dx when
y=(tan x)^cot x + (cot x)^tan x
5 years ago

Answers : (1)

Rinkoo Gupta
askIITians Faculty
80 Points
							

let y1=(tanx)cotx

taking log on both sides, we get

logy1=cotx.logtanx

diff.w.r.to x , get

(1/y1)dy1/dx=cotx.(1/tanx).sec2x+logtanx.(-cosec2x)

dy1/dx=(tanx)cotx .cosec2x[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)(-cosec2x)+logcotx.sec2x

= (cotx)tanx (sec2x)[-1+logcotx]

Hence dy/dx==(tanx)cotx .cosec2x[1- logtanx]+ (cotx)tanx (sec2x)[-1+logcotx]

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty

5 years ago
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