Differentiation

NZ Level 7 (NZC) Level 2 (NCEA)

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.

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

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

Find $f\left(x+h\right)$

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

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

`f`(`x`+`h`)−`f`(`x`)`h`.Hence find $f'\left(x\right)$

`f`′(`x`) by evaluating $\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$lim`h`→0(`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

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.

$\lim_{h\to0}\left(\frac{-6\left(-2+h\right)^2-\left(-6\times\left(-2\right)^2\right)}{h}\right)$lim`h`→0(−6(−2+`h`)2−(−6×(−2)2)`h`)

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

$\lim_{h\to0}\left(\frac{-6\left(-2+h\right)^2+24}{h}\right)$lim`h`→0(−6(−2+`h`)2+24`h`)

Above we have simplified the last term.

$\lim_{h\to0}\left(\frac{-6\left(4-4h+h^2\right)+24}{h}\right)$lim`h`→0(−6(4−4`h`+`h`2)+24`h`)

Above we expanded the brackets.

$\lim_{h\to0}\left(\frac{-24+24h-6h^2+24}{h}\right)$lim`h`→0(−24+24`h`−6`h`2+24`h`)

And again. Now simplifying we're left with:

$\lim_{h\to0}\left(\frac{24h-6h^2}{h}\right)$lim`h`→0(24`h`−6`h`2`h`)

Factorising out the $h$`h` we get:

$\lim_{h\to0}\left(\frac{h\left(24-6h\right)}{h}\right)$lim`h`→0(`h`(24−6`h`)`h`)

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

$\lim_{h\to0}\left(24-6h\right)$lim`h`→0(24−6`h`)

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

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