The idea here in ‘geometric’ terms is that in some vague sense a
curved line can be approximated by a straight line tangent to it. Of
course, this approximation is only good at all ‘near’ the point of
tangency, and so on. So the only formula here is secretly the formula
for the tangent line to the graph of a function. There is some hassle
due to the fact that there are so many different choices of symbols to
*write* it.

We can write some formulas: Let $f$ be a
function, and fix a point $x_o$. The idea is that *for $x$ ‘near’
$x_o$ we have an ‘approximate’ equality*
$$f(x)\approx f(x_o)+f'(x_o)(x-x_o)$$
We do *not* attempt to clarify what *either* ‘near’ or
‘approximate’ mean in this context. What is really true here is that
for a given value $x$, the quantity
$$f(x_o)+f'(x_o)(x-x_o)$$
is *exactly* the $y$-coordinate of the line *tangent* to the
graph at $x_o$.

The aproximation statement has many paraphrases in varying choices of symbols, and a person needs to be able to recognize all of them. For example, one of the more traditional paraphrases, which introduces some slightly silly but oh-so-traditional notation, is the following one. We might also say that $y$ is a function of $x$ given by $y=f(x)$. Let $$\Delta x = \hbox{ small change in $x$}$$ $$\Delta y= \hbox{ corresponding change in $y$ }=f(x+\Delta x)-f(x)$$

Then the assertion is that $$\Delta y\approx f'(x)\,\Delta x$$

Sometimes some texts introduce the following questionable
(but traditionally popular!) notation:
$$dy = f'(x)\;dx = \hbox{ approximation to change in $y$}$$
$$dx = \Delta x$$
and call the $dx$ and $dy$ *‘differentials’*. And then this whole
procedure is **‘approximation by differentials’**. A not
particularly enlightening paraphrase, using the previous notation, is
$$dy\approx \Delta y.$$
Even though you may see people writing this, don't do it.

More paraphrases, with varying symbols: $$f(x+\Delta x)\approx f(x) + f'(x)\Delta x$$ $$f(x+\delta)\approx f(x) + f'(x)\delta$$ $$f(x+h)\approx f(x) + f'(x)h$$ $$f(x+\Delta x)-f(x)\approx f'(x)\Delta x$$ $$y+\Delta y\approx f(x)+f'(x)\Delta x$$ $$\Delta y\approx f'(x)\Delta x$$

*A little history:* Until just $20$ or $30$ years ago,
calculators were not widely available, and especially not typically
able to evaluate trigonometric, exponential, and logarithm
functions. In that context, the kind of vague and unreliable
‘approximation’ furnished by ‘differentials’ was certainly worthwhile
in many situations.

By contrast, now that pretty sophisticated calculators are widely available, some things that once seemed sensible are no longer. For example, a very traditional type of question is to ‘approximate $\sqrt{10}$ by differentials’. A reasonable contemporary response would be to simply punch in ‘$1,0,\sqrt{}$’ on your calculator and get the answer immediately to 10 decimal places. But this was possible only relatively recently.

#### Example 1

For example let's approximate $\sqrt{17}$ by
differentials. For this problem to make sense at all *imagine that
you have no calculator*. We take $f(x)=\sqrt{x}=x^{1/2}$. *The idea
here is that we can easily evaluate ‘by hand’ both $f$ and $f'$
at the point $x=16$ which is ‘near’ $17$*. (Here
$f'(x)={1\over 2}x^{-1/2}$). Thus, here
$$\Delta x=17-16=1$$
and
$$\sqrt{17}=f(17)\approx f(16)+f'(16)\Delta
x=\sqrt{16}+{1\over 2}{ 1 \over \sqrt{16 }}\cdot 1=4+{1\over 8}$$

#### Example 2

Similarly, if we wanted to approximate $\sqrt{18}$ ‘by differentials’, we'd again take $f(x)=\sqrt{x}=x^{1/2}$. Still we imagine that we are doing this ‘by hand’, and then of course we can ‘easily evaluate’ the function $f$ and its derivative $f'$ at the point $x=16$ which is ‘near’ $18$. Thus, here $$\Delta x=18-16=2$$ and $$\sqrt{18}=f(18)\approx f(16)+f'(16)\Delta x=\sqrt{16}+{1\over 2}{ 1 \over \sqrt{16 }}\cdot 2=4+{1\over 4}.$$

Why not use the ‘good’ point $25$ as the ‘nearby’ point to
find $\sqrt{18}$? Well, in broad terms, the further away your ‘good’
point is, the worse the approximation will be. Yes, it is true that we
have little idea how good or bad the approximation is *anyway*.

It is somewhat more sensible to *not* use this idea for
numerical work, but rather to say things like $$\sqrt{x+1}\approx
\sqrt{x}+{1\over 2}{ 1 \over \sqrt{x }}$$ and $$\sqrt{x+h}\approx
\sqrt{x}+{1\over 2}{ 1 \over \sqrt{x }}\cdot h.$$ This kind of assertion
is more than any particular numerical example would give, because it
gives a *relationship*, telling how much the *output* changes
for given change in *input*, depending what *regime*
(=interval) the input is generally in. In this example, we can make
the *qualitative* observation that *as $x$ increases the
difference $\sqrt{x+1}-\sqrt{x}$ decreases*.

#### Example 3

Another numerical example: Approximate $\sin\,31^o$ ‘by
differentials’. Again, the point is *not* to hit $3,1,\sin$ on
your calculator (after switching to degrees), but rather to *
imagine that you have no calculator*. And we are supposed to remember
from pre-calculator days the ‘special angles’ and the values of trig
functions at them: $\sin \,30^o={1\over 2}$ and $\cos
\,30^o={\sqrt{3} \over 2}$. So we'd use the function $f(x)=\sin\,x$,
and we'd imagine that we can evaluate $f$ and $f'$ easily by hand at
$30^o$. Then
$$\Delta x=31^o-30^o=1^o=1^o\cdot {2\pi\hbox{ radians } \over 360^o}=
{2\pi\over 360}\hbox{ radians }$$
We have to rewrite things in radians
since we really only can compute derivatives of trig functions in
radians. Yes, this is a complication in our supposed ‘computation by
hand’. Anyway, we have
$$\sin\,31^o=f(31^o)=f(30^o)+f'(30^o)\Delta
x =\sin\,30^o+\cos\,30^o\cdot {2\pi\over 360}$$
$$={1\over 2}+{\sqrt{3} \over 2} {2\pi\over 360}$$ Evidently we are
to *also* imagine that we *know* or can easily *find*
$\sqrt{3}$ (by differentials?) as well as a value of $\pi$. *Yes*,
this is a lot of trouble in comparison to just punching the
buttons, and from a contemporary perspective may seem senseless.

#### Example 4

Approximate $\ln(x+2)$ ‘by differentials’, in terms of
$\ln x$ and $x$: This *non-numerical* question is somewhat more
sensible. Take $f(x)=\ln\,x$, so that $f'(x)={1\over x}$. Then
$$\Delta x=(x+2)-x=2$$
and by the formulas above
$$\ln(x+2)=f(x+2)\approx f(x)+f'(x)\cdot 2=\ln\,x+{2\over x}.$$

#### Example 5

Approximate $\ln\,(e+2)$ in terms of differentials: Use $f(x)=\ln\,x$ again, so $f'(x)={1\over x}$. We probably have to imagine that we can ‘easily evaluate’ both $\ln\,x$ and ${1\over x}$ at $x=e$. (Do we know a numerical approximation to $e$?). Now $$\Delta x = (e+2)-e = 2$$ so we have $$\ln(e+2)=f(e+2)\approx f(e)+f'(e)\cdot 2=\ln\,e+{2\over e}=1+{2\over e}$$ since $\ln\,e=1$.

#### Exercises

- Approximate $\sqrt{101}$ ‘by differentials’ in terms of $\sqrt{100}=10$.
- Approximate $\sqrt{x+1}$ ‘by differentials’, in terms of $\sqrt{x}$.
- Granting that ${d\over dx}\ln x={1\over x}$, approximate $\ln(x+1)$ ‘by differentials’, in terms of $\ln x$ and $x$.
- Granting that ${d\over dx}e^x=e^x$, approximate $e^{x+1}$ in terms of $e^x$.
- Granting that ${d\over dx}\cos x=-\sin x$, approximate $\cos(x+1)$ in terms of $\cos\,x$ and $\sin\,x$.