Differentiation

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

First Principles

Lesson

As we just saw, we can find the derivative (rate of change at a point), using first principles.

This formula 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)

Our next step is to see how to use the formula.

is the formula we use, so we need only know the function f(x) and then we can find the derivative.

Note how at this point, we don't need the point, we are not finding the derivative as just one point, we are finding the derivative at ALL the points. That is, we are finding the derivative function. It is a function that we can then use to find the derivative at any particular point we want.

Let's look at a simple example. The linear function $y=2x$`y`=2`x`. (we already know that this function has a gradient of $2$2, so that is the answer we are hoping for)

$=$= | |||

$=$= | |||

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$=$= | evaluating the limit is easy, as there are no h's left | ||

$=$= | $2$2 |

Concept question

Go now and complete the same process but for the function $f(x)=2x+5$`f`(`x`)=2`x`+5 and $f(x)=2x-3$`f`(`x`)=2`x`−3.

- What do you notice?
- Why is this the case?

_{Try this first, then check here for the answer.}

Here are some hints about using the first principles method

- The function $f(x)$
`f`(`x`) should always cancel out completely on the numerator. If it hasn't you will have made an error and you should go back and check. - The $h$
`h`on the denominator should always cancel out. Sometimes directly, like in this example, or sometimes through a factorisation and cancellation. If it doesn't, then you have made a mistake so go back and check your work.

Let's look at another example.

Let's find the derivative of the function $f(x)=3x^2+4x-5$`f`(`x`)=3`x`2+4`x`−5

Before we jump straight into the formula, let's first look at what f(x+h) is.

$f(x)=3x^2+4x-5$`f`(`x`)=3`x`2+4`x`−5 so,

$f(x+h)=3(x+h)^2+4(x+h)-5$`f`(`x`+`h`)=3(`x`+`h`)2+4(`x`+`h`)−5

$f(x+h)=3(x^2+h^2+2xh)+4x+4h-5$`f`(`x`+`h`)=3(`x`2+`h`2+2`x``h`)+4`x`+4`h`−5

$f(x+h)=3x^2+3h^2+6xh+4x+4h-5$`f`(`x`+`h`)=3`x`2+3`h`2+6`x``h`+4`x`+4`h`−5

$f(x+h)=3x^2+4x-5+3h^2+6xh+4h$`f`(`x`+`h`)=3`x`2+4`x`−5+3`h`2+6`x``h`+4`h`

$=$= | ||

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$=$= | ||

$=$= | $6x+4$6x+4 |

This gives us the capacity to find the gradient at any point on the curve $f(x)=3x^2+4x-5$`f`(`x`)=3`x`2+4`x`−5 by using the function $\frac{dy}{dx}=6x+4$`d``y``d``x`=6`x`+4.

So the gradient of the tangent at the point where $x=4$`x`=4, is $6\times4+4=28$6×4+4=28 (that's pretty steep!), and the gradient of the tangent at the point where $x=-0.5$`x`=−0.5 is $6\times(-0.5)+4=-3+4=1$6×(−0.5)+4=−3+4=1.

We can confirm by looking at the graph of the function.

Concept question

Go now and complete the same process but for the function $f(x)=2x+5$`f`(`x`)=2`x`+5 and $f(x)=2x-3$`f`(`x`)=2`x`−3.

- What do you notice?
- Why is this the case?

for $f(x)=2x+5$`f`(`x`)=2`x`+5

$f(x+h)=2(x+h)+5$`f`(`x`+`h`)=2(`x`+`h`)+5

so

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}\left(\frac{\left(2\left(x+h\right)+5\right)-\left(2x+5\right)}{h}\right)$limh→0((2(x+h)+5)−(2x+5)h) |

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}\left(\frac{\left(2x+2h+5\right)-\left(2x+5\right)}{h}\right)$limh→0((2x+2h+5)−(2x+5)h) |

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}\left(\frac{2x+2h+5-2x-5}{h}\right)$limh→0(2x+2h+5−2x−5h) |

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}\left(\frac{2h}{h}\right)$limh→0(2hh) |

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}2$limh→02 |

$\frac{dy}{dx}$dydx |
$=$= | $2$2 |

for $f(x)=2x-3$`f`(`x`)=2`x`−3

$f(x+h)=2(x+h)-3$`f`(`x`+`h`)=2(`x`+`h`)−3

so

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}\left(\frac{\left(2\left(x+h\right)-3\right)-\left(2x-3\right)}{h}\right)$limh→0((2(x+h)−3)−(2x−3)h) |

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}\left(\frac{\left(2x+2h-3\right)-\left(2x-3\right)}{h}\right)$limh→0((2x+2h−3)−(2x−3)h) |

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}\left(\frac{2x+2h-3-2x+3}{h}\right)$limh→0(2x+2h−3−2x+3h) |

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}\left(\frac{2h}{h}\right)$limh→0(2hh) |

$\frac{dy}{dx}$dydx |
$=$= | $\lim_{h\to0}2$limh→02 |

$\frac{dy}{dx}$dydx |
$=$= | $2$2 |

So what have you noticed?

That all three functions, $f(x)=2x,f(x)=2x+5$`f`(`x`)=2`x`,`f`(`x`)=2`x`+5 and $f(x)=2x-3$`f`(`x`)=2`x`−3 all have the same derviative, $\frac{dy}{dx}=2$`d``y``d``x`=2.

Can you see why this is?

Try drawing the graph of the $3$3 functions, then see what you notice.

Yep, they are all parallel. The constant terms of $+5$+5, and $-3$−3, simply moved the line $y=2x$`y`=2`x` up and down the plane. They had no change on the slope, and hence no effect on the gradient function.

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