Math Insight

Linear approximations: approximation by differentials


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 $\ln\,e=1$.


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