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Differentials

```
Date: 09/18/98 at 01:09:03
From: Maria sanabria
Subject: Differentials

I have to reach this conclusion:

If you can get the differentials of a function, you can differentiate
it, but if you can differentiate it, you can not necessarily get its
differentials.

```

```
Date: 09/18/98 at 07:35:27
From: Doctor Jerry
Subject: Re: Differentials

Hi Maria,

The standard definition of the differential of a real-valued function
f of a real variable is:

At a given point x, the differential df_x (df sub x; usually the x
is omitted) of f is the linear function defined on R by:

df_x(h) = f'(x) * h

Everyday usage of the differential often suppresses the fact that the
differential is a linear function. For example, if y = f(x) = x^2,
then we write:

dy = df = 2x * dx

where dx is used instead of h. This is for good reason. The finite
numbers dy and dx appearing in dy = 2x * dx can be manipulated to
obtain:

dy/dx = 2x

I feel that I haven't replied directly to your question. I think that
this is because I don't fully understand your question.

- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 09/19/98 at 13:57:17
From: Maria Stella Sanabria
Subject: Differentials

Thanks for your answer. I know that the question is a little bit
confusing, and at the beginning I thought it was a problem of the
little, so I am going to try to rephrase it.

What is the difference between finding the derivatives of a function
(dy/dx), and finding its differentials (dy, dx)?

In the books I've seen they define differentials supposing that f(x)
is differentiable.

My teacher gave a hint to reach this conclusion: if you can find the
differentials of f, then f is differentiable, but if f is
differentiable you can't necessarily find its differentials.

That is why I can prove this, starting with a function that is
differentiable.

Thanks.
```

```
Date: 09/22/98 at 08:53:36
From: Doctor Jerry
Subject: Re: Differentials

Hi Maria,

Suppose f(x) = x^2. To find the derivative of f we use the definition
of derivative: f'(x) is the limit as h->0 of the quotient

f(x+h) - f(x)
-------------
h

For this function, f'(x) = 2x.

Okay, this much is clear; there is no possible ambiguity.

The differential of f at x is defined to be the linear function df,
which is defined on all of R by:

df(h) = f'(x) * h

Often, the notation df(h) is shortened to df or, if y = f(x), then we
write dy instead of df. Then the above definition is:

dy = f'(x)*dx    or

dy/dx = f'(x)

Unless you are studying differential geometry, in which dx is
interpreted slightly differently, dx is not the differential of a
function. It is a variable, the same as h.

I think the definitions you found in the books and what I said above
are comparable. However, in elementary calculus the definitions change
a bit when one goes from functions of one variable to functions of two
or more variables. In one variable it doesn't matter whether one
defines the derivative of a function first, as I did above, and then
defines the differential (the linear function idea) or does the
reverse (see below). They are equivalent. For functions of more than
one variable, however, big differences become evident.

Here is the definition of differentiability that works for functions
of one or more variables.

Suppose that f is a function on (a,b) and p is a point of (a,b). Then
f is differentiable at p if a real number P can be found for which

|f(p+h)-f(p)-h*P|/|h| -> 0 as h -> 0

Suppose that f is differentiable at p, P is the derivative of f at p
and we write f'(p) = P. Also, the function L(h) = h*P, from R into R,
is called the differential of f at p, often written as df. The
differential is a mapping; the derivative is a number.

This is equivalent to the earlier definition of derivative, that is,
if one assumes either, the other follows.

Now, this form carries over to functions of two (and more) variables.
Suppose that f is a function defined on the set I of all (x,y) for
which a < x < y and c < y < d and (p,q) is a point of I. Then f is
differentiable at (p,q) if real numbers P and Q can be found for
which:

|f(p+h,q+k)-f(p,q)-h*P-k*Q|/sqrt(h^2+k^2) -> 0 as (h,k)->(0,0)

If f is differentiable, then f has partial derivatives at (p,q), but
the converse is not true.

- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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