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Question

If y=x-1 log x-1-x+1 log x+1, prove that dydc=log x-11+x


Solution

We have, y=x-1 logx-1-x+1 logx+1
Differentiating with respect to x,
dydx=ddxx-1 logx-1-x+1 logx+1        =x-1ddxlogx-1+logx-1ddxx-1-x+1ddxlogx+1+logx+1ddxx+1                     =x-1×1x-1ddxx-1+logx-1×1-x+1×1x+1×ddxx+1+logx+11        =1+logx-1-1+logx+1        =logx-1-logx+1        =logx-1x+1      So, dydx=logx-1x+1              

Mathematics
RD Sharma XII Vol 1 (2015)
Standard XII

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