Implicit Differentiation

**paile** · May 27th 2009, 07:14 AM

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?

**Soroban** · May 27th 2009, 08:35 AM

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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