Find dy /dx if x=ln (y-e ^-y)?



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Posted on (Answer #1)

You need to write y in terms of x, such that:

`x = ln (y - e ^(-y)) => y - e ^(-y) = e^x`

You need to differentiate both sides with respect to x, such that:

`(dy)/(dx) + e^(-y)*(dy)/(dx) = e^x`

Factoring out `(dy)/(dx)` yields:

`(dy)/(dx)(1 + e^(-y)) = e^x => (dy)/(dx) = (e^x)/(1 + e^(-y))`

Replacing ` y - e ^(-y)` for `e^x` yields:

`(dy)/(dx) = (y - e ^(-y))/(1 + e^(-y))`

Hence, evaluating `(dy)/(dx)` , using implicit differentiation, under the given conditions, yields `(dy)/(dx) = (y - e ^(-y))/(1 + e^(-y)).`

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