NZ Level 7 (NZC) Level 2 (NCEA)
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Rate of change functions
Lesson

With our work on Rates of Change we know that by applying the limiting chord process at various points on the curve $f(x)$f(x), we end up with a whole new function which we call the gradient function and which we denote by $f'(x)$f(x).

Instead of applying the limiting chord process to a number of points and doing a heap of calculations, we can fast track this whole process by completing the limiting chord process for any general point on the curve. We also have a name for this process, and it's called Differentiating Using First Principles.

Finding our Rate of Change Function

Let's recap the process of using first principles with the following question and video solution.

Worked Example

We want to find the derivative of $f\left(x\right)=-6x^2$f(x)=6x2 from first principles.

  1. Find $f\left(x+h\right)$f(x+h) in expanded form.

  2. Find $f\left(x+h\right)-f\left(x\right)$f(x+h)f(x).

  3. Find $\frac{f\left(x+h\right)-f\left(x\right)}{h}$f(x+h)f(x)h.

  4. Hence find $f'\left(x\right)$f(x) by evaluating $\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$limh0(f(x+h)f(x)h).

If we now wanted to find the gradient at a particular point on the curve, at say $x=-2$x=2, we can substitute this value into $f'(x)$f(x).

$f'\left(-2\right)=-12\times-2$f(2)=12×2

$f'\left(-2\right)=24$f(2)=24

Finding the gradient at a point using First Principles

The other way we could go about the above problem is b directly substituting $x=-2$x=2 into our first principles expression first, and then complete the limiting chord process. Let's take a look at doing it this way.

$\lim_{h\to0}\left(\frac{-6\left(-2+h\right)^2-\left(-6\times\left(-2\right)^2\right)}{h}\right)$limh0(6(2+h)2(6×(2)2)h)

Above we have replaced $x$x with $-2$2.

$\lim_{h\to0}\left(\frac{-6\left(-2+h\right)^2+24}{h}\right)$limh0(6(2+h)2+24h)

Above we have simplified the last term.

$\lim_{h\to0}\left(\frac{-6\left(4-4h+h^2\right)+24}{h}\right)$limh0(6(44h+h2)+24h)

Above we expanded the brackets.

$\lim_{h\to0}\left(\frac{-24+24h-6h^2+24}{h}\right)$limh0(24+24h6h2+24h)

And again. Now simplifying we're left with:

$\lim_{h\to0}\left(\frac{24h-6h^2}{h}\right)$limh0(24h6h2h)

Factorising out the $h$h we get:

$\lim_{h\to0}\left(\frac{h\left(24-6h\right)}{h}\right)$limh0(h(246h)h)

And now we cancel out the $h$h and are left with:

$\lim_{h\to0}\left(24-6h\right)$limh0(246h)

And so, substituting in $h=0$h=0 we get a gradient of $24$24. Just like we did above.

 

Outcomes

M7-10

Apply differentiation and anti-differentiation techniques to polynomials

91262

Apply calculus methods in solving problems

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