Differentiation 2. The METRIC Project, Imperial College. Imperial College of Science Technology and Medicine, 1996.

Transcription

1 Differentiation 2 The METRIC Project, Imperial College. Imperial College of Science Technology and Medicine,

2 Launch Mathematica. Type <<Mathetic`diffpack` Instructions for Getting Started hold down the shift key, and press the return key. Wait for Mathematica s response. (Note: be sure to use the ` symbol rather than the '. You may need to hunt for it on your keyboard: on most, it s in the top left corner.) This essential first step sets up Mathematica for this module. If you omit this bit, the special commands (see below) will not work. Mathematica Commands The following Mathematica commands may be useful to you in this module. Commands that come with Mathematica: D, Sin, Cos, Tan, Log, Together, Plot, Solve Special commands for this module: GiveQuestion, LastAnswer, PlotChord, PlotWithSP, ImplicitPlot For information on, say, the Solve command, type?solve hold down shift, and press return. 2

3 Experiment 1: Products and quotients Preparatory reading If you've performed the experiments in the Differentiation 1 module, you'll already know the importance of being able to find gradients of curved graphs, and you'll already know how to do this how to differentiate in the case of a fairly wide range of simple functions. We now seek techniques that allow us to differentiate things like y = x5 sin x + e cos x 2x + x 3, more complicated functions built from the simple ones we can already handle. In this experiment, you are asked to seek techniques for differentiating products and quotients. Thus, if we know how to differentiate f(x) and g(x), then we can also differentiate both f(x) g(x) and f(x)/g(x). In this experiment, you make use of Mathematica s D command. Take a moment to remind yourself what this command does. Try, for example D[Sin[x], x] D[x^3, x] etc. Don't forget to shift-return at the end of each line. 1) Try the following input lines: D[E^x Sin[x], x] D[Sin[x] Log[x], x] D[x^2 Cos[x], x] Investigate further. 2) Using your results from part 1, write a brief description of a rule for differentiating products of functions. 3) Is the output from D[u[x] v[x], x] in accord with your rule? Note down this output. 4) Use Mathematica to investigate differentiation of functions like sin x e x and ln x 2 (that is, of one x function divided by another). Briefly describe a rule for this kind of problem, and compare your rule with the output from the command D[u[x] / v[x], x] Note: the command Together places expressions over a common denominator. If you're already familiar with a rule for differentiating quotients and it doesn't match what Mathematica gives you, try 3

4 Together[ D[u[x] / v[x], x] ] instead. 5) Use the rule you have just discovered to differentiate tan x and sec x. Check using Mathematica. Post-experiment reading We can sum up the rules for differentiating products and quotients in the form of two formulae. The first, the product rule, is this: or d dx ( uv) = v du dx + u dv dx, [ uv] ' = u' v + uv'. The second, the quotient rule, can be stated like this: d u = 1 du dx v v dx u dv v 2 dx. This corresponds to the way Mathematica expresses the formula, but it is not the usual or preferred way of writing it. The quotient rule is more often expressed (and more easily remembered) as d u = dx v v du dx u dv dx v 2, or u ' v = u' v uv' v 2. Experiment 2: Composite functions Preparatory reading A composite function is one made from two or more simpler functions strung together. For example: y = e cos x x cosine exponential function y or 4

5 y = cos( e x ) x exponential function cosine y In this experiment, we seek a rule which enables us to differentiate a composite function when we know how to differentiate the two simple functions which make it up. 1) Try the following input lines: D[Sin[x^3], x] D[Cos[E^x], x] D[Tan[Log[x]], x] Investigate further. 2) Using your results from part 1, write a brief description of a rule for differentiating composite functions. 3) Is the output from D[u[v[x]], x] in accord with your rule? Note down this output. 4) Use the rule you have just discovered to differentiate (5 + x 2 ) 4. Check using Mathematica. Post-experiment reading The rule you have just discovered is called the chain rule. It can be written like this: if the derivative of u(x) is u'(x) and the derivative of v(x) is v'(x) then the derivative of u[v(x)] is v'(x) u' [v(x)]; or, perhaps more simply, like this: dy dx = dy du du dx. For example: if y = sin (5 + 11x) then we let u = x. Then y = cos u, which gives It follows that dy dx dy du = cosu and du dx = 11. = 11cosu =11cos ( 5+11x ). 5

6 Practice Questions We ve included a feature which allows you to get Mathematica to generate practice questions and their answers. There are three sets of questions on differentiation rules. To generate a question of the first kind, type GiveQuestion["product rule"] not forgetting to shift-return. To generate the answer for checking, type LastAnswer["product rule"] You can do this as often as you want: the questions are randomly generated, and repetitions should be rare. [Note: you do not need to retype the commands. Simply click on what you have already typed, and then shift-return.] The other two sets of questions are accessed by the commands GiveQuestion["quotient rule"] and GiveQuestion["composite functions"] Note: this section uses this module s special functions. If they fail to work, try going back to the Instructions for Getting Started at the beginning. Experiment 3: Max and min Preparatory reading A local maximum on a graph is the top of a hill : a point higher than all points close to it. Note that it isn t necessarily the highest point attained: merely the highest in its immediate vicinity. Similarly a local minimum is the bottom of a valley : a point lower than all points close to it. 6

7 A general term meaning either maximum or minimum is turning point. Many real-life problems have to do with maximising and minimising quantities: an insight into the nature of maxima and minima is a useful thing to have. In this experiment, we use calculus to seek one. In this experiment, you are asked to use Mathematica to define a function, whose graph you are then asked to plot. If you wanted, for example, to define the function f(x) = 4 x 2 and then plot it between 2 and 6, you would type f[x_] = 4 - x^2 Plot[f[x], {x, -2, 6}] remembering to shift-return after each line. (Note, however, that this particular example does not appear in the experiment!) Later, you are asked to solve for x the equation f '(x) = 0. To do this, type Solve[f'[x]==0, x] 1) Define the function f(x) = x 30x 2 8x 3 + 3x 4, and plot it for 4 x 4. Inspect your graph and estimate the positions of the stationary points: the points where the gradient is zero. Note down these estimated positions. 2) Use the special command PlotWithSP to generate a graph of f(x) showing the stationary points type PlotWithSP[f[x], {x, -4, 4}] What is the significance of the numerical output? 3) Use Mathematica to find the x-coordinates of the stationary points, by solving for x the equation f '(x) = 0. Using a technique of your own devising, find also the y-coordinates. Check that they correspond to your findings from parts 1 and 2. 4) Note down which of the stationary points are maxima and which are minima. Evaluate f '' at each of the stationary points that is, the derivative of the derivative. (To evaluate f '' at x = 2, say, simply type f''[2]). Seek a relationship between these second derivative values and the nature of the stationary points. Write a brief description of this relationship. 7

8 5) Repeat for the curve f (x) = x2 + 4 x + 4 x ) Write a brief explanation of the relationship between second derivatives and the nature of stationary points. 7) Study the curves f (x) = x 30x 2 4x 3 + 3x 4, and f (x) = x + 50x 2 20x 3 5x 4 + 4x 5 in the same way. Refine or amend your description and your explanation if necessary. Note: this experiment uses this module's special functions. If they fail to work, try going back to the Instructions for Getting Started at the beginning. Post-experiment reading The stationary points on the graph of y = f(x) are those points where the graph is locally horizontal: where f '(x) = 0. All turning points are stationary points, but not all stationary points are turning points: it's possible for the gradient to be zero at a point which is neither a maximum nor a minimum. This third type of stationary point is called a point of stationary inflexion. At a stationary point P where the second derivative is positive, the gradient is zero and rising. This means the gradient must be negative to the left of P, and positive to the right of P: we have a minimum. 8

9 Similarly, a stationary point where the second derivative is negative is a maximum. A stationary point where the second derivative is zero can be either a maximum, or a minimum, or a point of horizontal inflexion. Practice Questions There are two sets of questions on stationary points, accessible by the commands GiveQuestion["stationary points"] and GiveQuestion["max, min, etc"] Use LastAnswer in the normal way for the answers. Experiment 4: Implicitly defined functions Preparatory reading So far, we've only met graphs of functions of the form y = f(x). However, it's entirely possible to specify a graph by means of an equation of which y is not the subject: an example is the circle equation x 2 + y 2 = 4. 9

10 Equations such as this are said to define graphs implicitly. The circle example also defines a function y = f(x) implicitly if we restrict its range to be either all positive or all negative (since every input can have only one output). Each point on the graph of an implicitly defined function has a gradient, so it must be possible to make sense of the idea of the derivative of y with respect to x. In this experiment, you seek a way of doing that. This experiment uses the special command ImplicitPlot. To generate the graph of the implicit equation x 2 + y 2 = 4, type ImplicitPlot[x^2 + y^2 == 4, {x, -3, 3}, {y, -3, 3}] remembering to shift-return. 1) Use ImplicitPlot to generate the graphs of the implicit equations x 2 + y 2 = 4, y 2 = x 2 (3 - x), x 2 arctan(x y 2 ) = e y and sin x sin y = 0.03x. 2) Type the command ImplicitPictures to call up three images. Use ImplicitPlot to generate these images for yourself feel free to work by trial and error. 3) What is the derivative of x 2 with respect to x? Check using Mathematica. What, therefore, is the derivative of y 2 with respect to x, if y depends on x? Check by typing D[y[x]^2, x] (Why do you think D[y^2, x] isn't correct?) 4) What do we get if we differentiate the circle equation x 2 + y 2 = 4 with respect to x on both sides? Check by typing 10

11 D[x^2 + y[x]^2 == 4, x] Note down this equation. Using Mathematica's Solve command or by hand, solve it with respect to y'(x), and hence write down dy/dx in terms of x and y for this implicit function. 5) Repeat for the equation y 2 = x 2 (3 - x), and for others of your own devising. Post-experiment reading Implicit equations give us, typically, expressions for dy/dx which involve both x and y (ordinary explicit equations give us dy/dx as a function of x only). To find such expressions, we simply differentiate the implicit equation term by term, remembering that d dx ( f ( y) ) = f ' ( y ) dy dx (a result which is quite easy to prove from the chain rule). For example: y 2 = x 2 ( 3 x) 2y dy = 6x 3x2 dx dy 6x 3x2 =. dx 2y Practice Questions There is one set of questions on implicit differentiation, accessible by the command GiveQuestion["implicit functions"] Use LastAnswer in the normal way for the answers. 11