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If `x^y=e^(x-y),`show that `(dy)/(dx)=(logx)/({log(x e)}^2)` |
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Answer» `x^y = e^(x-y)` Taking log both sides, `log(x^y) = log(e^(x-y))` `=>ylogx = (x-y)loge` As `log e = 1`, `=>ylogx+y = x` `=>y(1+logx) = x` `y = x/(1+logx)` `dy/dx = ((1+logx)1-x(1/x))/(1+logx)^2` `=>dy/dx = (1+logx-1)/(loge+logx)^2` `=>dy/dx = logx/(log(xe))^2` |
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