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Question

If xy=exy, then find dydx at x=1

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Solution

Applying log on both sides

y logx=xy.......(1) [logxm=mlogx]
ylogx+y=xy=x1+logx
When x=1 then y=11+log1
=11+0
y=1
Differentiating equation (1) wrt x
yx+log xdydx=1dydx [(UV)=UV+UV]
log xdydx+dydx=1yx
dydx(logx+1)=xyx
dydx=xyx(1+log x)
To find dydx at x=1:
[dydx]x=1=111(1+log1)=01(1+0)=01=0

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