# Taylors Series

## Main Question or Discussion Point

I was going through the derivation of the Taylors series in my book (Engineering Mathematics by Jaggi & Mathur), and there was one step that escaped me. They proved that the derivative of f(x+h) is the same wrt h and wrt (x+h). If someone could explain that, Id be really grateful.

## Answers and Replies

I was going through the derivation of the Taylors series in my book (Engineering Mathematics by Jaggi & Mathur), and there was one step that escaped me. They proved that the derivative of f(x+h) is the same wrt h and wrt (x+h). If someone could explain that, Id be really grateful.
Do you mean df(x+h)/dh = df(x+h)/d(x+h)? If so that just does not make any sense, you cannot say that. However, engineering books and physics books on math sometimes use erroneous manipulation which leads to correct results. I am one who does not understand their manipulations so I cannot help with that, but I can see right now that is invalid.

for the derivation of Taylor Series... just keep doing integration by part until you get bored with it.

HallsofIvy
Homework Helper
Kummer, why does df(x+h)/dh = df(x+h)/d(x+h) make no sense?

By the chain rule, df(x+h)/dh= f'(x+h) since the derivative of x+ h, wrt h, is 1. To do
df(x+h)/d(x+h) more easily, change the variable. Let y= x+ h so the problem becomes
df(y)/dy. Of course, that is f"(y)= f'(x+h).

Kummer, why does df(x+h)/dh = df(x+h)/d(x+h) make no sense?

By the chain rule, df(x+h)/dh= f'(x+h) since the derivative of x+ h, wrt h, is 1. To do
df(x+h)/d(x+h) more easily, change the variable. Let y= x+ h so the problem becomes
df(y)/dy. Of course, that is f"(y)= f'(x+h).
I see what you did.

I am just mentioning I do not know, usually, differential usage. For instance, I have seen books write d(x+1) I am not sure what this means. And I have seen dx^2, this is even more mysterious.

Gib Z
Homework Helper
I have seen books write d(x+1) I am not sure what this means. And I have seen dx^2, this is even more mysterious.
Its just like in a change of variables, say we have the integral $$\int \frac{log x}{x} dx$$ and we let u= log x, then integral becomes $$\int u du$$ which, we could just write differently as $$\int log x d(log x)$$.

HallsofIvy
Homework Helper
You can differentiate a function of x with respect to any function of x.
$$\frac{dg(x)}{df(x)}= \frac{dg(x)}{dx}\frac{dx}{df(x)}= \frac{dg(x)}{dx}\frac{1}{\frac{df}{dx}}$$
by the chain rule.

In particular,
$$\frac{dg(x)}{dx^2}= \frac{\frac{dg}{dx}}{2x}$$

You can differentiate a function of x with respect to any function of x.
$$\frac{dg(x)}{df(x)}= \frac{dg(x)}{dx}\frac{dx}{df(x)}= \frac{dg(x)}{dx}\frac{1}{\frac{df}{dx}}$$
by the chain rule.

In particular,
$$\frac{dg(x)}{dx^2}= \frac{\frac{dg}{dx}}{2x}$$
It's a classic situation where a student of analysis (when he's only part way through his course) will cry foul. He will claim that differentiation is defined as an operation to be done on *one* function, by the taking of a particular limit. Unfortunately (or fortunately, I would say), physics and engineers follow an *algebraic* tradition, over pure analysis. We tend to just perform algebraic manipulations, without particular regard over the validity, until they fail, and we learn that step is not valid in some context. It says a lot that the algebraic route can work well enough that whole generations of physicists and engineers have used it to great effect without ever learning the distinction. In fact, only recently (by mathematical standards), has the algebraic method gained more legitimacy: search for non-standard analysis, or synthetic differential geometry.

As maybe obvious, this happens to be one of my personal little nooks of fascination...

Kummer, why does df(x+h)/dh = df(x+h)/d(x+h) make no sense?

By the chain rule, df(x+h)/dh= f'(x+h) since the derivative of x+ h, wrt h, is 1. To do
df(x+h)/d(x+h) more easily, change the variable. Let y= x+ h so the problem becomes
df(y)/dy. Of course, that is f"(y)= f'(x+h).
Thank you. My exams just finished, sorry for taking so long to reply.