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Dy/dx if y=[1+1/x]^x |
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Answer» ong>Answer: LOG y=X*1/(1+1/x)*logx + (1+1/x) Step-by-step explanation: taking log on both sides we get log y= log(1+1/x)^x log y= XLOG(1+1/x) LOGY = x*d/dylog(1+1/x)+(1+1/x)logx log y=x*1/(1+1/x)*logx + (1+1/x)*1 log y=x*1/(1+1/x)*logx + (1+1/x) |
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