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File: Chain Rule Pdf 170546 | Mc Ty Chain 2009 1
thechainrule mc ty chain 2009 1 aspecial rule the chain rule exists for dierentiating a function of another function this unit illustrates this rule in order to master the techniques ...

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         TheChainRule
                                                                    mc-TY-chain-2009-1
         Aspecial rule, the chain rule, exists for differentiating a function of another function. This unit
         illustrates this rule.
         In order to master the techniques explained here it is vital that you undertake plenty of practice
         exercises so that they become second nature.
         After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
            • explain what is meant by a function of a function
            • state the chain rule
            • differentiate a function of a function
                                         Contents
          1. Introduction                                                          2
          2. A function of a function                                              2
          3. The chain rule                                                        2
          4. Some examples involving trigonometric functions                       4
          5. A simple technique for differentiating directly                        5
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          1. Introduction
          In this unit we learn how to differentiate a ‘function of a function’. We first explain what is
          meant by this term and then learn about the Chain Rule which is the technique used to perform
          the differentiation.
          2. A function of a function
                                   2
          Consider the expression cosx . Immediately we note that this is different from the straightforward
                                                          2
          cosine function, cosx. We are finding the cosine of x , not simply the cosine of x. We call such
          an expression a ‘function of a function’.
          Suppose, in general, that we have two functions, f(x) and g(x). Then
                                                y = f(g(x))
          is a function of a function. In our case, the function f is the cosine function and the function g
          is the square function. We could identify them more mathematically by saying that
                                                                2
                                         f(x) = cosx    g(x) = x
          so that
                                                      2        2
                                          f(g(x)) = f(x ) = cosx
          Now let’s have a look at another example. Suppose this time that f is the square function and
          g is the cosine function. That is,
                                                 2
                                         f(x) = x     g(x) = cosx
          then
                                        f(g(x)) = f(cosx) = (cosx)2
                              2      2         2
          Weoften write (cosx) as cos x. So cos x is also a function of a function.
          In the following section we learn how to differentiate such a function.
          3. The chain rule
          In order to differentiate a function of a function, y = f(g(x)), that is to find dy, we need to do
                                                                               dx
          two things:
          1.  Substitute u = g(x). This gives us
                                                 y = f(u)
          Next we need to use a formula that is known as the Chain Rule.
          2.  Chain Rule
                                              dy = dy × du
                                              dx    du   dx
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                                                 KeyPoint
           Chain rule:
           To differentiate y = f(g(x)), let u = g(x). Then y = f(u) and
                                            dy = dy × du
                                            dx    du   dx
         Example
                                              2
         Suppose we want to differentiate y = cosx .
                  2
         Let u = x so that y = cosu.
         It follows immediately that
                                       du =2x     dy = −sinu
                                       dx         du
         The chain rule says
                                            dy = dy × du
                                            dx   du   dx
         and so
                                         dy = −sinu×2x
                                         dx
                                                        2
                                             = −2xsinx
         Example
         Suppose we want to differentiate y = cos2x = (cosx)2.
         Let u = cosx so that y = u2
         It follows that
                                     du = −sinx         dy = 2u
                                     dx                 du
         Then
                                         dy = dy ×du
                                         dx     du   dx
                                             = 2u×−sinx
                                             = −2cosxsinx
         Example
                                                10
         Suppose we wish to differentiate y = (2x−5) .
         Now it might be tempting to say ‘surely we could just multiply out the brackets’. To multiply
         out the brackets would take a long time and there are lots of opportunities for making mistakes.
         So let us treat this as a function of a function.
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         Let u = 2x−5 so that y = u10. It follows that
                                     du = 2          dy = 10u9
                                     dx              du
         Then
                                         dy = dy ×du
                                         dx     du   dx
                                             = 10u9×2
                                                         9
                                             = 20(2x−5)
         4. Someexamplesinvolvingtrigonometricfunctions
         In this section we consider a trigonometric example and develop it further to a more general
         case.
         Example
         Suppose we wish to differentiate y = sin5x.
         Let u = 5x so that y = sinu. Differentiating
                                     du = 5         dy =cosu
                                     dx             du
         From the chain rule
                                          dy = dy ×du
                                          dx     du   dx
                                              = cosu×5
                                              = 5cos5x
         Notice how the 5 has appeared at the front, - and it does so because the derivative of 5x was 5.
         So the question is, could we do this with any number that appeared in front of the x, be it 5 or
         6 or 1, 0.5 or for that matter n ?
             2
         So let’s have a look at another example.
         Example
         Suppose we want to differentiate y = sinnx.
         Let u = nx so that y = sinu. Differentiating
                                   du = n             dy =cosu
                                   dx                 du
         Quoting the formula again:
                                           dy = dy × du
                                           dx   du   dx
         So
                                          dy = cosu×n
                                          dx
                                              = ncosnx
         So the n’s have behaved in exactly the same way that the 5’s behaved in the previous example.
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