$\displaystyle Let\ x=e^t.\ Show\ that:$

$\displaystyle {\frac{d^2y}{dx^2}}={\frac{1}{x^2}} \left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)$

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- Apr 24th 2010, 04:35 AMacevipaDifficult differentiation question
$\displaystyle Let\ x=e^t.\ Show\ that:$

$\displaystyle {\frac{d^2y}{dx^2}}={\frac{1}{x^2}} \left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)$ - Apr 24th 2010, 04:37 AMskeeter
- Apr 24th 2010, 04:38 AMProve It
- Apr 24th 2010, 04:41 AMacevipa
- Apr 24th 2010, 04:52 AMacevipa
It's just a "show" question. So could it have something to do with starting from $\displaystyle \frac{d^2y}{dx^2}$ and reducing it into a form which only has x and t. Not really sure, how you would do that though.

- Apr 24th 2010, 05:14 AMschmeling
this question is a pretty straight forward use of the chain rule in higher derivatives.

There is no need for you to know the relation between y and t because as you can see in the question this relation is expressed as a differential.

what you need for solving this is to notice that if x=e^t then t=ln(x)

then dt/dx= x^(-1) and the second derivative is d^2t/dx^2= -x^(-2)

now using the chain rule

d^2y/dt^2= (d^2y/dt^2) * (dt/dx)^2 + (dy/dt) * (d^2t/dx^2)

its enough that you substitute for the values of dt/dx and d^2t/dx^2 what you derived in the beginning and you're done.

All it takes is understanding that the chain rule can be used for higher derivatives as well.

- Apr 24th 2010, 05:21 AMacevipa
- Apr 24th 2010, 05:40 AMschmeling
this is the chain rule for a second derivative of a function. You either need to have a formula booklet for that or use the Faa di Bruno's formula to derive it yourself.

Its beyond my latex using ability to post the actual derivation so far (ive just joined today) so I assumed that they expect you to know the chain rule for the second derivative rather than derive it but if you want to find out more about how to derive that yourself just read up on Faa di Bruno.