# Thread: Probably quite straight forward...

1. ## Probably quite straight forward...

I haven't done mv calculus in a while ... I'm stuck trying to understand a line of a solution I've been given. For this question T is a constant and x and t are variables.

If $\displaystyle f'(x)=g(x)$

$\displaystyle \frac{\partial}{\partial x}f(\frac{x}{\sqrt{T-t}}) = \frac{1}{\sqrt{T-t}}g(\frac{x}{\sqrt{T-t}})$ - I'm OK with this

$\displaystyle \frac{\partial ^2}{\partial x^2}f(\frac{x}{\sqrt{T-t}}) = \frac{1}{T-t}g'(\frac{x}{\sqrt{T-t}})$ - And this

$\displaystyle = \frac{x}{(T-t)^{3/2}}g(\frac{-x}{\sqrt{T-t}})$ using $\displaystyle g'(x) = -xg(x)$

I can see how the last equality holds given the "using...", but I have no idea where that came from. Any help?

Possibly there is some bad notation going on with the first and last statements - it should be f'(x*) = g(x*) and g'(x*) = -x*g(x*) or something, maybe...

2. ## Derivatives

Hello RanDom
Originally Posted by RanDom
I haven't done mv calculus in a while ... I'm stuck trying to understand a line of a solution I've been given. For this question T is a constant and x and t are variables.

If $\displaystyle f'(x)=g(x)$

$\displaystyle \frac{\partial}{\partial x}f(\frac{x}{\sqrt{T-t}}) = \frac{1}{\sqrt{T-t}}g(\frac{x}{\sqrt{T-t}})$ - I'm OK with this

$\displaystyle \frac{\partial ^2}{\partial x^2}f(\frac{x}{\sqrt{T-t}}) = \frac{1}{T-t}g'(\frac{x}{\sqrt{T-t}})$ - And this

$\displaystyle = \frac{x}{(T-t)^{3/2}}g(\frac{-x}{\sqrt{T-t}})$ using $\displaystyle g'(x) = -xg(x)$

I can see how the last equality holds given the "using...", but I have no idea where that came from. Any help?

Possibly there is some bad notation going on with the first and last statements - it should be f'(x*) = g(x*) and g'(x*) = -x*g(x*) or something, maybe...
I have no idea where $\displaystyle g'(x) = -xg(x)$ comes from either. This will only be the case if $\displaystyle g(x)$ is of the form:

$\displaystyle g(x) = Ae^{-\tfrac{1}{2}x^2}$

Do you have anything to indicate that this might be the case?

3. Ohhhhhhh.... Yes I do! $\displaystyle f$ is the standard normal distribution hahahaha..... What an idiot.