# How can we neatly explain chain rule of differentiation

My students are often getting confused while using chain rule for complicated functions. For example

$$f(x)=\tan^3\left(\sqrt{x^2+x+1}\right)$$ Most of the students wrote $f'(x)$ wrongly as

$$f'(x)=3 \tan^2\left(\sqrt{x^2+x+1}\right) \times \frac{1}{2\sqrt{x^2+x+1}} \times (2x+1)$$

Can I have any comments on how better I can teach chain rule?

• This is interesting. I have not seen this kind of error before. It seems like they just missed out one step, but it is unusual that most of the students left out the same step. Feb 18 '16 at 6:34
• I wonder if they would have made the same mistake if the function had been written $\left(\tan\left(\sqrt{x^2+x+1}\right)\right)^3$.
Feb 18 '16 at 13:44
• Is this really a problem of misunderstanding the chain rule or a problem of being fooled by the notation convention used for powers of trig functions? Feb 18 '16 at 15:02
• This may be in the same category as the kind of mistake where a student is trying to differentiate $\sin\cos x$ and applies the product rule. For students operating at that level, it can be difficult to get them to articulate why they do what they do. They may not be thinking of a specific and incorrect interpretation of the notation; they may simply not be thinking at all.
– user507
Feb 20 '16 at 20:50

Here is one way to write the solution which cuts down on errors:

$$\frac{d}{dx} \tan^3(\sqrt{x^2+x+1}) = \frac{d(\tan^3(\sqrt{x^2+x+1}))}{d(\tan(\sqrt{x^2+x+1}))} \frac{d(\tan(\sqrt{x^2+x+1}))}{ d(\sqrt{x^2+x+1})} \frac{d\sqrt{x^2+x+1}}{d (x^2+x+1)} \frac{d(x^2+x+1)}{dx}$$

Forcing the student to set up all of the computations first, before performing them, seems to cut down on the number of errors. I think this helps to solve the problem of remembering which step you are on while you compute a given derivative.

• A few things: (1) Excellent suggestion (2) It might be useful to teach the students to use variable substitution to make it clear what to differentiate in each case. For instance, the first factor looks pretty confusing as written, but if you make the substitution $u = \tan\sqrt{x^2 + x + 1}$ it becomes quite obvious. (Of course you have to remember to reverse the substitution after taking the derivative.) (3) Mathematicians always complain about this, so it's probably worth saying that this is a memory aid, not a rigorous derivation of the chain rule. Feb 18 '16 at 8:15
• @David Z My personal objection to the $u$-formulas you find in certain elementary textbooks is that they seem to reinforce the student's idea that we need to memorize different chain rules for all the different elementary functions. Instead, I want them to think about one chain-rule that applies to all functions because I want them to understand more and memorize less. The notation of $u$ itself is not the objection (I think). Feb 18 '16 at 10:53
• @JamesS.Cook I'm not sure if I'm familiar with these $u$-formulas. Can you give an example? The only chain rule I know of is the standard one, $dy/dx = dy/du\times du/dx$ (for single-variable functions under appropriate conditions). Feb 18 '16 at 13:18
• @DavidZ Some calculus books will incorporate the chain rule into the statement of every formal rule of differentiation, for example writing $\frac{d}{dx} u^n = nu^{n-1} \frac{d u }{d x}$. Feb 18 '16 at 16:40
• Or, $\frac{d}{dx} \cos( u) = -\sin(u) \frac{du}{dx}$ or $\frac{d}{dx} \ln (u) = \frac{1}{u} \frac{du}{dx}$ and... Feb 18 '16 at 16:56

This is a slight variation on Steven Gubkin's suggestion, incorporating David Z's suggestion in his comment on that answer.

Suggest to the students that whenever they encounter a complicated nexted function, they explicitly write it as a chain of several simpler functions, e.g.: $$y=u^3$$ $$u=\tan(v)$$ $$v=\sqrt{w}$$ $$w=x^2+x+1$$

Then the chain rule says simply that the derivative of the first variable $y$ with respect to the last variable $x$ is the product of the derivaives of all of the individual links in the chain:

$$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv}\cdot \frac{dv}{dw}\cdot \frac{dw}{dx}$$

So $$\frac{dy}{dx} = 3u^2 \cdot \sec^2(v) \cdot \frac{1}{2\sqrt{w}} \cdot (2x+1)$$ and the problem can be completed by back-substituting in the expressions for each of the variables in terms of the simpler ones.

• It would also be good if this kind of skill was taught in precalculus, when students are learning function transformations. Feb 18 '16 at 22:43