Implicit differentiation

$x=e^u$

I can do dy/dx of this and I get
$\frac{dx}{dy}=e^u.\frac{du}{dx}$
and I invert it to get dy/dx.
However I'm getting lost trying to find $\frac{d^2y}{dx^2}$ of the above equation.any help will be appreciated

I don't know where you are getting y from.
You should be finding du/dx not dy/dx.
You could take logs of both sides to get rid of the need to differentiate implicitly.
Else do what you were doing

Have you covered c4

$x=e^u$

I can do dy/dx of this and I get
$\frac{dx}{dy}=e^u.\frac{du}{dx}$
and I invert it to get dy/dx.
However I'm getting lost trying to find $\frac{d^2y}{dx^2}$ of the above equation.any help will be appreciated

Yes.

It seems an odd equation to be differentiating wrt y. Is this the whole question?

I dont want to argue , but I need to find d^2y/dx^2 and finding du/dx wont get me there

$x^2\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+2y=0$

using the substituion, $x=e^u$, solve.

$x=e^u$

I can do dy/dx of this and I get
$\frac{dx}{dy}=e^u.\frac{du}{dy}$
and I invert it to get dy/dx.
However I'm getting lost trying to find $\frac{d^2y}{dx^2}$ of the above equation.any help will be appreciated

So, by the chain rule:

$\displaystyle\frac{dy}{du}=\frac{dy}{dx}\times \frac{dx}{du}$


$\displaystyle=\frac{dy}{dx}\times e^u$

$\displaystyle\frac{dy}{du}=x\frac{dy}{dx}$

and just substitute that in.

For the second derivative, using product rule and chain rule:

$\displaystyle\frac{d^2y}{du^2}=\frac{d^2y}{dx^2} \times \left(\frac{dx}{du}\right)^2+\frac{dy}{dx}\times \frac{d^2x}{du^2}$

Can you take it from there?

Edit: You don't actually need to find dy/dx, just the terms involving dy/dx, which in this case is x dy/dx. Similarly for d2y/dx2.

which gives me using the product rule:

$\frac{d^2y}{dx^2}=e^{-u}(-e^{-u}\frac{dy}{du}+e^{-u}\frac{d^2y}{du^2})[br]$
I should then able to tidy it up.would that be right?