1.

Dy/dx if y=[1+1/x]^x​

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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