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Modulus Sin(pi X ) issue.

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But wouldn't the chain rule in this case require evaluating the derivative of |x| at 0 (since 0^2 is still 0)? I get why this failing doesn't say anything about the limit existing thanks to your post, but how can you get around d/dg |g| at 0?

Thanks, that makes more sense

Recall the definition of a derivative:

\displaystyle \dfrac{df}{dx} = \lim_{\delta x \rightarrow 0} \dfrac{\delta f}{\delta x} = \lim_{\delta x \rightarrow 0} \dfrac{f(x+\delta x)-f(x)}{\delta x} \newline \newline

For this limit (and hence the derivative) to exist, we must find that the limit is the same when delta x approaches 0 from the positive or the negative direction so there is no ambiguity i.e. we must have

\displaystyle \lim_{\delta x \rightarrow 0^+} \dfrac{\delta f}{\delta x} = \lim_{\delta x \rightarrow 0^-} \dfrac{\delta f}{\delta x}

.

This is the problem that we have when dealing with the derivatives of moduli in general: these two limits are different*.

However, when we consider the function

f(u)=|u|, u \geq 0

i.e. when we limit the domain to the non-negative real numbers (as we have done implicitly with

u=x^2

in the example), we do not have the same ambiguity. The point we want to consider, at x=0 (and u=0) is the boundary of the domain, and it only makes sense to take the limit from the positive direction. In this case,

\displaystyle \dfrac{df}{du}=\lim_{\delta u\rightarrow 0} \dfrac{\delta f}{\delta u}=\lim_{\delta u \rightarrow 0^+} \dfrac{\delta f}{\delta u}

,

which is well defined. Hence the derivative exists.

In summary: if we define our function to be |x| over all real x, the derivative does not exist; if we define our function to be |x| over all non-negative x, the derivative does exist. I suppose the key thing to take home from this is that whether or not the derivative of a function exists depends upon the domain over which the function is defined.

*If you want to perform the calculation:

Spoiler

(edited 8 years ago)

Reply 21

username1162972

We coukd go into proper maths and technicalities. But think of it like this. at x=k where k is an interger we have zeroes right? now draw a tangent at that point. See what goes wrong?

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Modulus Sin(pi X ) issue.

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