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Question

If y = (x)tanx+(tanx)xthenfinddydx

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Solution

y=xtanx+(tanx)x
Let
u=xtanx and v=(tanx)x
Now,
u=xtanx
Taking log both sides, we get
logu=tanxlogx
Differentiating above equation w.r.t. x, we have
1ududx=tanx(1x)+logx(sec2x)
dudx=u(tanxx+sec2xlogx)
dudx=xtanx(tanxx+sec2xlogx)
Similarly,
v=(tanx)x
Taking log both sides, we have
logv=xlog(tanx)
1vdvdx=log(tanx)1+x(1tanxsec2x)
dvdx=v(log(tanx)+xsec2xtanx)
dvdx=(tanx)x(log(tanx)+xsec2xtanx)
Therefore,
dydx=dudx+dvdx
dydx=xtanx(tanxx+sec2xlogx)+(tanx)x(log(tanx)+xsec2xtanx)

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