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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    I came across this math problem,where i had to apply Implicit Differentiation to determine the derivative of y=x^e^-x^2.So in solving this problem,I introduced the natural log on either sides of the equation,to give Iny=Inx^e^-x^2,then appling one of the rules for logarithmic funtions,I made the exponent of my funtion to be the coefficient of Inx.Then i began differentating:

    y=x^e^-x^2
    Iny=Inx^e^-x^2
    =e^-x^2Inx
    1/y dy/dx=-2xe^-x^2Inx+e^-x^2/x .......................(A)
    I understand that I came to (A) through the use of the Chain Rule,the I understant that I had to make dy/dx the subject of my formula,so as to have:
    dy/dx=y(-2xe^-x^2Inx+e^-x^2/x)
    Therefore this is the part where I experiance a problem,I substitute y by its initial value (i.e x^e^-x^2) to get:

    dy/dx=x^e^-x^2(-2xe^-x^2Inx+e^-x^2/x) .........................(B).I arrived at (B) as my final solution,so I dont know if this is the correct solution then do I still need to go any further than I've come,do I need to proceed any further from my end point?
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  2. #2
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    Lexington, MA (USA)
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    Hello, paile!

    By the way, the natural log is written l\text{-}n, not I\text{-}n.


    \text{Differentiate: }\;y \:=\:x^{(e^{\text{-}x^2})}


    Take logs: . \ln y \:=\ln x^{(e^{\text{-}x^2})} \;=\; e^{\text{-}x^2}\!\cdot\!\ln x

    Differentiate implicitly: . \frac{1}{y}\,\frac{dy}{dx} \;=\;-2x\,e^{\text{-}x^2}\ln x + \frac{e^{\text{-}x^2}}{x}

    Then: . \frac{dt}{dx} \;=\;y\left[-2x\,e^{\text{-}x^2}\ln x + \frac{e^{\text{-}x^2}}{x}\right]

    Therefore: . \frac{dy}{dx}\;=\;x^{\left(e^{\text{-}x^2}\right)}\left[-2x\,e^{\text{-}x^2}\ln x + \frac{e^{-x^2}}{x}\right]<br />

    Do I still need to go any further?

    No . . . Your work is excellent!

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