`y = ln(e^-x + xe^-x)` Differentiate the function.

Textbook Question

Chapter 3, 3.6 - Problem 19 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

Posted on

`y=ln(e^-x+xe^-x)`

`y'=1/(e^-x+xe^-x) d/dx(e^-x+xe^-x)`

`y'=(1/(e^-x+xe^-x))(-e^-x+e^-x+x(-1)e^-x)`

`y'=(-xe^-x)/(e^-x+xe^-x)`

`y'=(-x)/(e^x(e^-x+xe^-x))`

`y'=(-x)/(e^0+xe^0)`

`y'=-x/(x+1)`

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hkj1385 | (Level 1) Assistant Educator

Posted on

`y = ln{(e^-x) + x*(e^-x)}`

taking antilog both sides we get

`e^y = (e^-x) + x*(e^-x)`

differentiating both sides

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

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

`or, dy/dx = {-x(e^-x)}/{(e^-x) + x*(e^-x)}`

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