# Differentiability and differential of a funtion

## Main Question or Discussion Point

For months I have been staring into this expression, and I cannot visualize what the hell omega represents...

f(x)-f(x0)=f'(x0)(x-x0)+$$\omega$$(x)*(x-x0)

Where $$\omega$$(x)(=$$\omega$$(x;$$\Delta$$x)) is a continuous function in point x0 and equals zero in that point

or lim, as x approaches x0 of omega(x)= omega(x0)=0

I do not completely understand this statement above. What does it represent? How do you understand this?

Thanks

Omega is the error function, in this sense: if you draw a function, then you can draw it's tangent line at a certain point x0. The distance between the tangent line and the function is given by $$\omega(x)(x-x_0)$$.

I don't think there is another way to visualize it. The omega isn't really all that important, it's more of a technical tool...

Hurkyl
Staff Emeritus
Gold Member
One way to think of it is that $\omega$ is what's leftover after "taking off" everything you know about f from the first-order analysis.

You know that, for example, the parabola y=x2 has a tangent line y=2x-1 at the point (1,1). Near that point, the parabola and the tangent line look similar. Have you ever then subtracted the two functions to better visualize how the parabola differs from the line? (if not, you should try it from time to time)

Dividing off the extra factor of (x-1) is just taking another step further.

If you're trying to understand the derivative of a function, then understanding $\omega$ isn't really important beyond the fact it converges to zero at x=x0.

On the other hand, if you are trying to analyze f(x) by saying it behaves similarly to its tangent line, then $\omega(x)$ becomes important in the sense that it's the part you have to show doesn't contribute to whatever phenomenon you're interested in.

One way to think of it is that $\omega$ is what's leftover after "taking off" everything you know about f from the first-order analysis.

You know that, for example, the parabola y=x2 has a tangent line y=2x-1 at the point (1,1). Near that point, the parabola and the tangent line look similar. Have you ever then subtracted the two functions to better visualize how the parabola differs from the line? (if not, you should try it from time to time)

Dividing off the extra factor of (x-1) is just taking another step further.

If you're trying to understand the derivative of a function, then understanding $\omega$ isn't really important beyond the fact it converges to zero at x=x0.

On the other hand, if you are trying to analyze f(x) by saying it behaves similarly to its tangent line, then $\omega(x)$ becomes important in the sense that it's the part you have to show doesn't contribute to whatever phenomenon you're interested in.

I understand derivatives quiet well. Differential of the function too. I understand whats the purpose of both. But this, what I wrote, is what defines function's differentiability. If omega is not 0, then the function is not differentiable, am I right? I am just trying to understand how come this omega is so powerful and what it does to a function, that defines its differentiability.

Omega is the error function, in this sense: if you draw a function, then you can draw it's tangent line at a certain point x0. The distance between the tangent line and the function is given by $$\omega(x)(x-x_0)$$.

I don't think there is another way to visualize it. The omega isn't really all that important, it's more of a technical tool...
Distance in what way? Is a picture possible?