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# Secondary Magazine - Issue 121: Sixth Sense

Created on 23 April 2015 by ncetm_administrator
Updated on 20 May 2015 by ncetm_administrator

# Sixth Sense

With past papers complete, exams underway, and students fully focussed on the here and now, this seems like a good time to ponder: “what shall we do with Year 12 when they come back after the modules?”

It has always struck me as unsatisfactory that during Year 12 students don’t make much progress – certainly not to halfway – on the overall A-level “calculus journey”, especially if they followed any GCSE or extension course that introduced differentiation in Year 11. They may well arrive in the sixth form knowing a rule for differentiating $\inline \dpi{80} \fn_jvn f\left ( x \right ) = kx\textup{\textup{}}^{n}$; hopefully (especially if they’re sitting C2!) they now know how to reverse this process, and they know that doing so helps them answer two seemingly-unconnected questions: “what did I differentiate to get this function I’m integrating?”, and “what’s the area enclosed between this curve and the x-axis?”

Take a deep breath and consider how much there is for your students to learn, understand and practise between now and next year’s A2 exams (depending on board, and whether or not they’re doing any Further Mathematics): the chain rule, the product rule, the quotient rule, calculus involving exponential and logarithmic function, calculus involving trigonometric functions, implicit differentiation, integration by substitution and by parts, solving differential equations, parametric differentiation, everything to do with hyperbolic functions, reduction formulae … And breathe out ...

So, it seems to make sense to try and make some headway between now and the end of term, and I will start by tackling the chain rule. However, I’m convinced that they’re not familiar enough with functional notation to cope with

$\dpi{80} \fn_jvn \small \left (fg \left ( x \right ) \right )' = f'\left ( g\left ( x \right ) \right ) \times g'\left ( x \right )$

nor fluent enough with the operator $\dpi{80} \fn_jvn \small \frac{d}{dx}$ to justify telling them that $\dpi{80} \fn_jvn \small \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$.

Rather than starting by asserting these arcane statements, let’s approach the problem head on, and find the formula for the gradient of the tangent for each of the following:

1  $\dpi{80} \fn_jvn \small y=\left ( 2x-3 \right )^{3}$
(by expanding, differentiating and factorising; students may need nudging in this direction)

2  $\dpi{80} \fn_jvn \small y=\left ( x^{2}+2 \right )^{3}$
(again, by first expanding).

Students can then claim and verify (using the same “expand, differentiate, factorise” technique) a prediction for the gradient formula for

3  $\dpi{80} \fn_jvn \small y = \left ( 4x+5 \right )^{3}$

and

4  $\dpi{80} \fn_jvn \small y = \left ( 3x^{3} +1\right )^{4}$

Having tried a few of these, students can believe with some confidence:

Prior knowledge:  $\dpi{80} \fn_jvn \small \frac{d}{dx}\left ( x^{n} \right )=nx^{n-1}$

New knowledge: $\dpi{80} \fn_jvn \small \frac{d}{dx}\left \left ( (function ^{n} \right \right ))$ $\dpi{80} \fn_jvn \small n\left ( function \right )^{n-1}\times function differentiated$

In my scheme of work I’ll now allow time for lots of practice: finding equation of tangents and normals, location and nature of stationary points and points of inflection, and, in particular, questions such as

What did I differentiate to get $\dpi{80} \fn_jvn \small 16x\left (x ^{2} -5\right )^{3}$?

Students will be encouraged to “guess and check”: I won’t say that a suggestion is “wrong”, instead my response will always be “fine, differentiate it and see what happens”. I’m aiming for this thought process (if not this layout):

Guess: $\dpi{80} \fn_jvn \small \left (x ^{2}-5 \right )^{4}$

Differentiate: $\dpi{80} \fn_jvn \small 4\left (x ^{2}-5 \right )^{3}\times 2x$

Thought: We need to multiply this by 2 to get the correct answer

Therefore: $\dpi{80} \fn_jvn \small \int 16x\left (x ^{2}-5 \right )^{3} dx = 2 \left (x^{2} -5 \right )^{4}$

There’s lots of practice needed here.

While the new “power of a function” rule beds in, but continuing the theme of gradient rules, next I will introduce the function $\dpi{80} \fn_jvn \small e^{x}$. I like doing this as an investigation, and there’s plenty of scope for IT use here too.

I anticipate that the conversation will go something like this:

Me: First, let's sketch $\dpi{80} \fn_jvn \small y=2^{x}$.What’s the gradient of the tangent to the curve $\dpi{80} \fn_jvn \small y=2^{x}$ at the point where $\dpi{80} \fn_jvn \small x=0$?

Student A: Differentiate to get $\dpi{80} \fn_jvn \small x\times 2^{x-1}$ and put in 0.

Student B: No, because then the gradient would be 0 and it isn’t: look at my sketch.

Student A: Does our rule only work if the power is a number, not an x?

Me: Let’s draw a picture [using geogebra, desmos, or similar], draw a tangent and estimate the gradient that way.

Student C: Ok, I get 0.7.

Student D: Yes, so do I.

Me: Why are we so trusting of the software? How else could we estimate the gradient of the tangent?

Student D: We could draw some chords and work out their gradients.

Me: Let’s do so.

Student B: Can we use a spreadsheet?

Me: Good plan.

Student B: Done it. I’m getting 0.7 too.

Me: Ok. What’s the gradient of the tangent to the curve $\dpi{80} \fn_jvn \small y=3^{x}$ at the point where $\dpi{80} \fn_jvn \small x=0$?

Student E: Computer says 1.1.

Student B: Excel does too.

Me: So what can we deduce from these two results?

Students will discuss, ponder and – fingers crossed – suggest the following conjecture (though this may require some steering!):

Conjecture: It’s reasonable to assume that there’s a number between 2 and 3 which has the property “the gradient of the tangent to the curve $\dpi{80} \fn_jvn \small y = \left ( that number \right )^{x}$ at $\dpi{80} \fn_jvn \small x=0$ is equal to 1”

Me: Ok, let’s hunt down that number!

And then, in a flurry of graphing software and / or Excel-exploring, the first few digits of e are discovered.

Me: Right, now let’s find the gradient of the tangent to this new-found curve $\dpi{80} \fn_jvn \small y = 2.718^x{}$ at $\dpi{80} \fn_jvn \small x=1$$\dpi{80} \fn_jvn \small x=2$ etc. What do you notice?

Me: In other words …

Students: Ah ha! $\dpi{80} \fn_jvn \small \frac{d}{dx}\left ( e^{x} \right ) = e^{x}$

(I’ve possibly moved into a world where students say exactly what I hope they will say, so will stop there).

Next, we will investigate (supported by IT) gradient rules for (a)  $\dpi{80} \fn_jvn \small y=4e^{x}$ (b)  $\dpi{80} \fn_jvn \small y=e^{5x}$ (c)  $\dpi{80} \fn_jvn \small y=e^{x^{2}}$.

where claims will be made and verified, revising in (a) transformations of graphs and in (b) first rewriting as $\dpi{80} \fn_jvn \small y=\left (e ^{x} \right )^{5}$ and then using the recently-learned rule for $\dpi{80} \fn_jvn \small y=\left ( function \right )^{n}$.

(c) is starting to generalise this rule, from “power of a function” to “function of a function”, and it seems to be saying that $\dpi{80} \fn_jvn \small \frac{d}{dx}\left ( e^{function} \right ) = \left ( e^{function} \right )\times function differentiated$

I will support this by revisiting the earlier “power of a function” rule, and helping the students to think of $\dpi{80} \fn_jvn \small \left ( 3x-1 \right )^{3}$ as $\dpi{80} \fn_jvn \small cube\left ( 3x-1 \right )$, where the function $\dpi{80} \fn_jvn \small cube \left ( x \right )$ differentiates to $\dpi{80} \fn_jvn \small 3\times square\left ( x \right )$. Then we will read our “power of a function” rule as a “function of a function” rule – and one that is consistent with the claim just made.

You can imagine similar explorative lessons that look at $\dpi{80} \fn_jvn \small \dpi{80} \fn_jvn \small ln x$ and $\dpi{80} \fn_jvn \small ln \left ( function \right )$, and $\dpi{80} \fn_jvn \small \cos x$ and $\dpi{80} \fn_jvn \small \cos \left ( function \right )$, $\dpi{80} \fn_jvn \small \sin x$ and $\dpi{80} \fn_jvn \small \sin function$, etc. Tempting as it will be to present a shortcut to a formal statement of the chain rule, developing numerous “function of a function” rules over time makes it far more likely that the students will develop both conceptual understanding and procedural fluency, and they will conclude for themselves that

If we know that $\dpi{80} \fn_jvn \small \frac{d}{dx}\left (f \left ( x \right ) \right )={f}'\left ( x \right )$ then it follows that $\dpi{80} \fn_jvn \small \frac{d}{dx}\left (f \left ( function \right ) \right )={f}'\left ( function \right )\times function differentiated$

When that comes from them rather than from me, I know that the route we’ve taken along this stretch of the calculus journey, though not the shortest, has certainly been the right one.

John Partridge is Assistant Head at Kings College London Maths School. He runs much of the school’s outreach and CPD activities, including a programme for teachers new to teaching further Pure modules. Further information can be found here.

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