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P1 Chapter 12 & 15

CIE Centre A-level Pure Maths. P1 Chapter 12 & 15. © Adam Gibson. Extending differentiation. In Chapter 11 we studied composite functions. Or a “function of a function”. A simple example:. What is . ?. What is . ?. How about . ?. BORING!. The chain rule - Introduction. g. f.

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P1 Chapter 12 & 15

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  1. CIE Centre A-level Pure Maths P1 Chapter 12 & 15 © Adam Gibson

  2. Extending differentiation In Chapter 11 we studied composite functions Or a “function of a function” A simple example: What is ? What is ? How about ? BORING!

  3. The chain rule - Introduction g f Let’s add 1 to x so we can find the gradient gradient=2 gradient=3 g f Of course this example is easy; but it shows that: the rate of change of f(g(x)) with respect to x is equal to the rate of change of g with respect to x multiplied by the rate of change of f with respect to g

  4. The chain rule - continued Mathematically: “The Chain Rule” This looks obvious, but remember You cannot do this: See p. 183 and p. 184 for a careful discussion of the derivation of the chain rule

  5. The chain rule – Examples Let’s go back to our first example Use the chain rule: Much easier than using the binomial theorem!! 

  6. The chain rule – Examples This rule allows to differentiate more complicated functions. Here’s an example: What is … ? Express f(x) as a composite function. We know: and:

  7. The chain rule – Examples Now we can apply the chain rule: Q: What are the stationary points of f ? The chain rule is very important in real world applications of calculus, because we are often dealing with several quantities (e.g. weather - time, temperature, pressure, density), many of which vary depending on each other.

  8. A real world example Coffee Remember our coffee cup? Question : how fast is the level of the coffee rising? h r

  9. A real world example - continued It can be shown (using integration) that the volume above the half way point varies with the height as follows: The cup is already half full. If we also know that the coffee is being poured at a rate of 42 cm3 per second, then at what rate will the height be increasing when h = 1cm? The question asks us to find The Chain Rule tells us:

  10. Second derivatives (Chapter 15) Let’s start by looking at our old friend, the quadratic curve: Sketch the curve on paper, and sketch the derivative on the same axes

  11. Second derivatives continued What is the equation of the derivative? Now draw the derivative of the derivative

  12. Second derivatives – definitions and notation The derivative of the derivative is usually called the second derivative It’s written like this: or sometimes Don’t get confused about this. We don’t write: or What we are really doing is applying the operator twice; it’s similar to a composite function.

  13. Second derivatives continued Sketch the first and second derivatives of:

  14. So far we have learned … line(ar) The first derivative of a quadratic is a : constant The second derivative of a quadratic is a : quadratic The first derivative of a cubic is a : line(ar) The second derivative of a cubic is a : The second derivative of a quartic (x4) is - quadratic This pattern should be easy to understand before we continue – remember,

  15. A little quiz Look at these curves. Which is the “odd one out”? B C A E D F

  16. Curvature The answer is D – all other curves have negative curvature Curvature is measured by the second derivative. Look: • The gradient is always increasing • The second derivative is always positive • The curvature is positive • The gradient is always decreasing • The second derivative is always negative • The curvature is negative

  17. Stationary points revisited We can use the idea of the second derivative to improve our method of identifying maxima and minima. Let’s return to the cubic equation we looked at earlier: The method given on p. 102 of the textbook is absolutely accurate, but can be slow and awkward – because we must evaluate the first derivative at 2 points. Look at the graph again, and think about curvature…

  18. Stationary points revisited Therefore: zero gradient and positive curvature => minimum while zero gradient and negative curvature => maximum

  19. The calculation in detail at a stationary point which is positive for x>1, negative for x<1 Hence: is a minimum, is a maximum (Don’t forget to calculate the function value!)

  20. Follow the right procedure! Make sure to read and understand the beginning of Section 15.3 p. 230. Not all examples are as easy as the one I’ve just presented! • In particular, think carefully about: • What if f(x) is undefined? • What if f’(x) is undefined? • What if f’’(x) is zero? The last case, f’’(x) or is particularly tricky. Please remember, f(x) has a stationary point at x=x0

  21. Zero curvature and inflexion points An inflexion point is a point about which the curvature of the function changes sign have any inflexion points? Does Note – stationary points are not always inflexion points, it depends on the curvature or second derivative. Consider the function y=x3 at x=0. < 0 So we see that the curvature changes sign; first it bends one way, then the other way. This is a Point of Inflexion = 0 > 0

  22. Inflexion points continued What can we say about the tangent at inflexion points? Answer – the tangent crosses the curve! But this only applies to smooth, differentiable functions. Let’s look at some “nastier” curves: First derivative not defined here; but is it an inflexion point? Yes! Note: this type will almost certainly not be on the P1 exam. First derivative not defined here; but is it an inflexion point? No!

  23. Higher order derivatives So far we have looked at first and second derivatives, which are very commonly used in a variety of applications. higher order derivatives are also sometimes useful. it is easy to understand how to find them; just use the same procedures. Called “third, fourth, fifth derivative” etc. Example: find the fourth derivative of

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