![]() ![]() So, how does the chain rule apply to 'y = sin(4x)'? Well, firstly differentiate the sine part of the equation and that is simply cosine. You start with the outermost function and work your way in. The way one decides which function is to be differentiated first is a bit like selecting which of the sets of cutlery to use at a very formal dinner. All the chain rule simply states is that you differentiate the first function and multiply the newly obtained derivative by the derivative of the second function. (The letter 'u' was chosen randomly to allow for the easy expression of the first function). In the example above 'y = sin(4x)', the two functions are 'sin(u)' and '4x' where 'u = 4x'. The chain rule is used when there is the need to differentiate a two function composite. When the sine curve is differentiated it almost magically transforms into a cosine curve which is a wonderful example of how maths works! If you'd like to see diagrammatic representations of the sine curve and the cosine curve then the internet will not disappoint. This pattern alternates until your patience runs out and you can't be bothered to draw the graph anymore! Mathematically of course it can repeat until infinity. Once a 180 degree cycle has passed where the curve rises to 'y = 1' then the next 180 degree cycle will show the curve going down to 'y = -1'. This is repeated every 180 degrees but there is one difference. If one has a graph with a curve 'y = sin(x)' then a sine wave will be seen which is a curve that starts at the origin, rises to 'y = 1' at the point 'x = 90 degrees' (or in radians, pi/2) and then sinks to 'y = 0' again at 'x = 180 degrees' (or pi radians). If you find a source which shows you the proof behind the knowledge that the derivative of 'sin(x)' is 'cos(x)' then it is well worth having a look! When first learning the list of derivatives of trigonometric functions one could be forgiven for thinking that the subject is a little tedious but the mathematics behind the list is beautiful. If the gradient is 39 then you would expect to see a steep incline of the curve at that point. ![]() Thus for the curve 'y = x^3 + 2x^2 + 4' you can obtain the derivative 'dy/dx = 3x^2 + 4x' and the gradient at the point 'x = 3' is: Then if you wish to find out the gradient at the point on the curve where 'x' equalled say 3, you just put 'x = 3' into the derivative. This new expression is known as the derivative and it is the derivative of 'y' in this case, with respect to 'x'. With a curve it isn't quite as simple to obtain the gradient because you can't just take the value of 'm' from the equation as it doesn't exist! In order to obtain the gradient at a specific point on a curve you first have to differentiate the equation of the curve as is shown in the question. If the graph showed a straight line then it would be possible to get an equation of the form 'y = mx + c' where 'm' is the constant gradient. As the line on the graph is a curve, the gradient will be different for different values of 'x'. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |