# Chain Rule

Hello:

These are not h/w problems but something from the class notes which I am not able to fully understand. I have two questions stated below

Question1
We are doing chain rule and function of several variables. To explain the prof has first explained about single variables and then gone to composite functions and vector functions. But since I cannot understand single variables, no point in going to other topics

Anyway the class note is as follows:

The differentiability of f at xo is equivalent to
( f(xo+h) - (f$$^{'}$$(x0)h) ) / h -> 0 i.e. the function f is approximated to first order by function x-> f(xo) + f$$^{'}$$(x0)(x-xo) (where x = xo + h)

The function h-> f(xo) + f$$^{'}$$(x0)h is a function of the form: a constant function (f(xo) + a linear map (f$$^{'}$$(x0)h). Such a function is called an affine function. Thus differentiability can be formulated as the statement: there is a linear map $$\phi$$: R->R such that the affine function f(xo) + $$\phi$$(x-xo) is a first-order approximation to f at xo.

My problem is as follows
- it is the last line which states that differentiability can be formulated by the linear map. Can someone explain what the notes are trying to say. I have been on this for about 3 days and not able to grasp it

Question 2:
Not entirely related to this, but assuming you have a function f(x1,x2,x3) = x1x2x3 + x1*x3. How do I prove that it is continuously differentiable. If I prove that partial derivatives exist, will it prove that they are continuously differentiable.

Thanks

Asif

Related Calculus and Beyond Homework Help News on Phys.org
Gib Z
Homework Helper
Q1. It means if a function f(x) is differentiable at a point $x_0$, the linear function that best approximates f(x) at x_0 is $$g(x) = f(x_0) + f' (x_0) (x - x_0)$$ ie its tangent.

To say the differentiability of f at x_0 is equivalent to it being approximated by that tangent is somewhat of a circular argument however, as the function must be differentiable at x_0 to obtain f'(x_0) .

Q2) What do you mean by "continuously differentiable"? If you mean repeatedly or infinitely differentiable, then showing the first partial derivatives exist only shows that they are once differentiable. You must show all further partials also exist.

EDIT: Ahh from Hall's reply, it appears the question was referring to whether or not the partials are continuous functions in themselves. My bad for misinterpreting.

Last edited:
HallsofIvy