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If `x=e^y+e^((y+ tooo))`, where `x >0,t h e nfin d(dy)/(dx)`A. `(1+x)/(x)`B. `(1)/(x)`C. `(1-x)/(x)`D. `(x)/(1+x)` |
Answer» Correct Answer - C We have, `x=e^(y+e^(y+e^(y+..." to "oo)))` `implies" "x=e^(y+x)` `implies" "log_(e)x=(y+x)` `implies" "(1)/(x)=(dy)/(dx)+1" "["Differentiating with respect to x"]` `implies" "(dy)/(dx)=(1-x)/(x)` |
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