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Question

If (tan1x)y+ycotx=1 , then find dydx

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Solution

Let y=(tan1x)y+ycotx1
Now let u=(tan1x)y and v=ycotx
starting with u, we get,
u=(tan1x)y
logu=ylog(tan1x)
1u.dudx=yddxlog(tan1)+log(tan1)dydx
1u.dudx=y(tan1x).11+x2+log(tan1x)dydx
dudx=y(tan1x)y(tan1x).11+x2+(tan1x)y.log(tan1x)dydx
dudx=y.(tan1x)y11x2+(tan1x)y.log(tan1x)dydx ------ ( 1 )
And now v=ycotx
Taking log we get,
logv=cotxlogy
On differentiating we get,
1vdvdx=cotxddxlogy+logyddxcotx
1vdvdx=cotxydydxcosec2x.logy
dvdx=ycotx.cotxy.dydxycotx.cosec2x.logy
dvdx=ycotx1.cotx.dydxycotx.cosec2x.logy ----- ( 2 )
Now, dydx=dudx+dvdx0

dydx=y.(tan1x)y11+x2+(tan1x)y.log(tan1x)dydx+ycotx1.cotx.dydxycotx.cosec2x.logy

dydx(tan1)y.log(tan1x)dydxycotx1.cotx.dydx=y(tan1x)y11+x2ycotx.cosec2x.logy

dydx(1(tan1x)y.log(tan1x)ycotx1.cotx)=y.(tan1x)y11+x2ycotx.cosec2x.logy
dydx=y(tan1x)y11+x2ycotx.cosec2x.logy(1(tan1x)y.log(tan1x)ycotx1.cotx





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