Feeling a little bit more confident about my calculus skills I was hoping to check if this is correct. Let’s say you have:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]g(|f(x)|)[/tex]

And you want to take the derivative with respect to x, well using the chain rule you get:

[tex]\bigl( g(|f(x)|) \bigr)' = (|f(x)|)' g'(|f(x)|)[/tex]

Looking more closly at [itex](|f(x)|)'[/itex]. Defining it as:

[tex]\left( +\sqrt{(f(x))^2} \right)'[/tex]

Then using the chain rule we get:

[tex](|f(x)|)' = 2f(x) f'(x) \: \frac{1}{2} \: \frac{1}{ +\sqrt{(f(x))^2} }[/tex]

Simplifying:

[tex](|f(x)|)' = f'(x) \frac{f(x)}{|f(x)|}[/tex]

Substituting back in the original equation and sorting out the problem of when f(x) = 0:

[tex]\bigl( g(|f(x)|) \bigr)' = \begin{cases}

f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) & \text{if $ f(x) \neq 0$} \\ \\

\lim_{f(x) \rightarrow 0} \left(f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right) & \text{if $ f(x) = 0$ and $ \lim_{f(x) \uparrow 0} \left( f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right) = \lim_{f(x) \downarrow 0} \left( f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right)$} \\ \\

\text{Undefinied} & \text{if $ f(x) = 0$ and $ \lim_{f(x) \uparrow 0} \left( f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right) \neq \lim_{f(x) \downarrow 0} \left( f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right)$}

\end{cases}

[/tex]

Correct? I know it all seems a bit over the top (especially when you look at the tex for it ) but I like things to be well defined and in a form like this where I understand it better.

Edit: I made a mistake, it should be correct now

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# Derivitave of a Modulus

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