Note for Chain Rule - CR by Placement Factory
- Chain Rule - CR
- Note
- Quantitative Aptitude
- Placement Preparation
- 234 Views
- 33 Offline Downloads
- Uploaded 1 year ago
Text from page-1
The Chain Rule mc-TY-chain-2009-1 A special 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 1 c
Text from page-2
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 Consider the expression cos x2 . Immediately we note that this is different from the straightforward cosine function, cos x. We are finding the cosine of x2 , 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 g(x) = x2 f (x) = cos x so that f (g(x)) = f (x2 ) = cos x2 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, f (x) = x2 g(x) = cos x then f (g(x)) = f (cos x) = (cos x)2 We often write (cos x)2 as cos2 x. So cos2 x is also a function of a function. In the following section we learn how to differentiate such a function. 3. The chain rule dy In order to differentiate a function of a function, y = f (g(x)), that is to find , 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 2 c
Text from page-3
Key Point Chain rule: To differentiate y = f (g(x)), let u = g(x). Then y = f (u) and dy dy du = × dx du dx Example Suppose we want to differentiate y = cos x2 . Let u = x2 so that y = cos u. It follows immediately that The chain rule says dy = − sin u du du = 2x dx dy du dy = × dx du dx and so dy = − sin u × 2x dx = −2x sin x2 Example Suppose we want to differentiate y = cos2 x = (cos x)2 . Let u = cos x so that y = u2 It follows that Then du = − sin x dx dy = 2u du dy dy du = × dx du dx = 2u × − sin x = −2 cos x sin x Example Suppose we wish to differentiate y = (2x − 5)10 . 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. 3 c
Text from page-4
Let u = 2x − 5 so that y = u10 . It follows that du =2 dx Then dy = 10u9 du dy dy du = × dx du dx = 10u9 × 2 = 20(2x − 5)9 4. Some examples involving trigonometric functions In this section we consider a trigonometric example and develop it further to a more general case. Example Suppose we wish to differentiate y = sin 5x. Let u = 5x so that y = sin u. Differentiating dy = cos u du du =5 dx From the chain rule dy du dy = × dx du dx = cos u × 5 = 5 cos 5x 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 12 , 0.5 or for that matter n ? So let’s have a look at another example. Example Suppose we want to differentiate y = sin nx. Let u = nx so that y = sin u. Differentiating du =n dx Quoting the formula again: dy = cos u du dy dy du = × dx du dx So dy = cos u × n dx = n cos nx So the n’s have behaved in exactly the same way that the 5’s behaved in the previous example. 4 c
Leave your Comments