My book is telling me

d/du((e^u)dy/dx)

=(e^u)dy/dx + (e^u)(d2y/dx2).(dx/du)

I understand where the first part has come from, but not sure how the second part was derived. Would anyone kindly provide me with an explanation?

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- Mar 21st 2009, 07:41 AMErghhhSolving differential equations using a change of variable
My book is telling me

d/du((e^u)dy/dx)

=(e^u)dy/dx + (e^u)(d2y/dx2).(dx/du)

I understand where the first part has come from, but not sure how the second part was derived. Would anyone kindly provide me with an explanation? - Mar 21st 2009, 08:43 AMADARSHErghhhhhhh....:p
This will make it clear

$\displaystyle \frac{d}{du}\{e^u\frac{dy}{dx}\} $

$\displaystyle = \frac{dy}{dx}\frac{d}{du}(e^u) +e^u \frac{d}{du}(\frac{dy}{dx}) $

Differnetiation of dy/dx wrt u will be given by (using chain rule)

$\displaystyle = \frac{y'}{dx} \cdot \frac{du}{dx}$

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So differnetiation will be

$\displaystyle =\frac{dy}{dx}e^u + e^u \frac{d^2(y)}{dx^2} \cdot \frac{dx}{du} $