New Zealand
Level 7 - NCEA Level 2

# 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