NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Limit and First Principle (investigation) LIVE
Lesson

It's time now to combine a number of ideas together. 

  • Average rate of change
  • Gradient of a line
  • Instantaneous rate of change

By combining these three things we are going to arrive at an understanding for the gradient at a point.

Have a play with this applet.  There is one fixed point $P$P, and another point $Q$Q.  You can move $Q$Q towards $P$P.  

  • As you do this, watch what happens to the average rate of change between $P$P and $Q$Q, remembering that the average rate of change is the slope of the line between $P$P and $Q$Q?  
  • If I wanted to know the instantaneous rate of change at point $P$P, how can we estimate this using the points  $P$P and $Q$Q?
  • Where is the best place to position $Q$Q to make this estimate?

What you should have found is that the best way to estimate the instantaneous rate of change is to bring $Q$Q as close as possible to $P$P.  This makes the gradient of the line between $P$P and $Q$Q, really close to the gradient of the tangent at $P$P

Getting a little more formal

Let me take some time to set up the above diagram with you.  I'd suggest you draw this in your book as you go.

Start with a set of axes, and the curve $y=f(x)$y=f(x).  Any curve will work, but maybe for now make it a similar shape to mine.  

Now, we select a point $P$P on the curve. It will have an $x$x coordinate of $x$x, so the point $P$P has coordinates $(x,f(x))$(x,f(x)).

Let's suppose I want to find the gradient at Point $P$P.  (we know already that this is reflected in the gradient of the tangent at point $P$P.)

Now, let's add a point $Q$Q, that is also on the curve $f(x)$f(x), and is some distance $h$h units away from $P$P (in the $x$x direction).  Then we can mark on $Q$Q, the distance $h$h and the coordinate of $Q$Q, $(x+h,f(x+h))$(x+h,f(x+h)) 

Now, to find the gradient of the tangent at $P$P, we could use the methods of coordinate geometry, (finding the rise and run), but doing this for many different points on the curve can get tedius.  Instead, we are going to estimate the gradient of the tangent at $P$P, using the section $PQ$PQ.  

I can hear you screaming at me. "But wait, that isn't a very good estimation, $PQ$PQ looks nothing like the tangent!"   At the moment, you are correct, just wait and see though, as I introduce our idea of limits in a minute and we will get those lines a lot closer.

So let's use the gradient of the section $PQ$PQ.

Gradient is found using $\frac{rise}{run}$riserun, which we can also write as $\frac{\text{change in y}}{\text{change in x}}$change in ychange in x

We can use the greek letter delta $(\Delta$(Δ) to indicate the words "change in", so our gradient is now

$m=\frac{\Delta y}{\Delta x}$m=ΔyΔx

We can see from our diagram that

$m=\frac{\Delta y}{\Delta x}=\frac{g}{h}$m=ΔyΔx=gh

Also, using the coordinates of P and Q, our change in $y$y ($\Delta y$Δy) is $f(x+h)-f(x)$f(x+h)f(x) and our change in $x$x ($\Delta x$Δx) is $(x+h)-x$(x+h)x so 

$m$m $=$= $\frac{\Delta y}{\Delta x}$ΔyΔx
$\frac{g}{h}$gh $=$= $\frac{f(x+h)-f(x)}{(x+h)-x}$f(x+h)f(x)(x+h)x
$=$= $\frac{f(x+h)-f(x)}{h}$f(x+h)f(x)h

 

All these statements are therefore equivalent

$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}=\frac{f(x+h)-f(x)}{h}$m=ΔyΔx=y2y1x2x1=f(x+h)f(x)h

 

I haven't forgotten... our current $PQ$PQ is not very similar to the tangent at $P$P.  Here is where the mathmagic happens! 

We move $Q$Q, closer to $P$P.  

Closer, and closer still

And even closer, until you can barely recognise that the distance h exists. 

Mathematically we describe this imagining of h (the distance between $P$P and $Q$Q) as getting closer and closer to $0$0 as a limit.  We write $h\rightarrow0$h0

First Principles

The process we have just undertaken has resulted in an understanding (and also usefully) has resulted in a formula to find the gradient at a point $P$P on a curve $f(x)$f(x).  

This formula is what is required when we are finding the derivative (which is the name we now use for the slope) from first principles (which basically means from scratch).  It gives us the instantaneous rate of change at point $P(x,f(x)$P(x,f(x) on any curve $f(x)$f(x).  

(We will talk more about the notations for the derivative a bit later) 

 

 

 

 

 

Outcomes

M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

91578

Apply differentiation methods in solving problems

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