Differentiation-JAKE DEACON

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1 Differentiation-JAKE DEACON Differentiation is the method of finding the gradient formulae (or the derivative). When we are given any equation we can differentiate said equation and find the gradient formulae and eventually the gradient. In algebraic terms an equation is. We then differentiate this and put in the form which is the same as. When differentiated = n. This may seem confusing but it is the same as MULTIPLY BY THE POWER, THEN TAKE 1 AWAY FROM THE POWER. Example: = 3-3 has become 3(multiply by power) (tak e 1 off the power) 3 has become 3(multiply by power) or no x (take 1 off the power) 3 - the same as 3. Question: =(6-7

2 Fractional and surd differentiation For fractional and surd differentiation we treat the equation like we have the previous equations. Example: y= / this the same as because you put coefficient of x as numerator and the power as negative power to the power in the denominator = (do not forget to minus the power even though you would initially think the answer would be ) Surd and fractional differentiation can also intertwine. When faced with a fractional power we have to break it down. The numerator of power is the same as the root, and the denominator of a power is the same as the power. Surd example: Y= =2 ( the power multiplies outside the surd ) Example: Y= / x = /

3 Differentiation given a point When we re given a point it s very easy, we just insert the value of x into the Example: = 3-3 At x=2 = (3x4)-3 = 9 form of the equation. And to find an equation when given a point We work out the form giving us the gradient. After this we have all parts needed to form the equation y-y1=m(x-x1) Example: at co-ordinate (2,10) = 10x-7 At x=2 = 20-7=13 y-y1=m(x-x1) y-10=13(x-2) y=13x-16 we can also use differentiation to find the equation of a normal(a line perpendicular to an equation or point). For these graphs we use the negative reciprocal of the tangent for the gradient and work out the equation from there using the same method as above.

4 Revision Notes Differentiation y = x n dy/dx = nx n-1 Examples (different levels of difficulty) y = x 4 dy/dx = 4x 3 y = 2x 4 dy/dx = 8x 3 y = x 5 + 2x -3 dy/dx = 5x 4-6x -4 Formula to work out Y or a line y-y_1=m(x-x_1) Where m= The gradient of the line Min and Max points Minimum points are negative to positive Maximum points are positive to negative

5 THE SECOND DERIVATIVE THE QUICK AND SIMPLE GUIDE By Richard Gibbs

6 What is it? For this we need to know what the derivative is this is when you differentiate the equation of the line: dy/dx=nx^n-1 e.g. Y=x^3 dy/dx=3x^2 We have used this many times in recent lessons and is a key feature in working out the gradients of curved lines, but has a range of uses

7 What is it? This means that the second derivative is when you differentiate the derivative. So for example if the equation of the line was y=x^3 you would have to first differentiate this to give you the derivative This would become dy/dx=3x^2 You then have to differentiate this This would then become 6x

8 How is it written? The second derivative can be written in two different ways: The first being d^2y/dx^2 The second being f (x)=6x Below shows the answers for the example on the previous slide:

9 Why is it used? The second derivative can be used as an easier way of determining the nature of stationary points, e.g. Whether they are maximum points, minimum points or points of inflection A stationary point occurs when the gradient is 0. This means we can use the derivative to work out what x is and therefore y You can then substitute x into the second derivative to give you the type of stationary point...

10 Key Information!!!!... If d 2 y is positive, then it is a minimum point dx 2 If d 2 y is negative, then it is a maximum point dx 2 If d 2 y = zero, then it could be a maximum, minimum d 2 y or point of inflection (If d 2 y/dx 2 = 0, you must test the values of dy/dx either side of the stationary point)

11 Example for y=x 3-27x At stationary points, dy/dx = 0 and dy/dx = 3x 2-27 If this is equal to zero, 3x 2-27 = 0 Hence x 2-9 = 0 (dividing by 3) So (x + 3)(x - 3) = 0 So x = 3 or -3 d 2 y/dx 2 = 6x When x = 3, d 2 y/dx 2 = 18, which is positive. When x = -3, d 2 y/dx 2 = -18, which is negative. there is a minimum point at x=3 and a maximum point at x=-3

12 Maths Revision Differentiation Maths Revision Differentiation Matthew Halls, Y12 Differentiation is the process by which you can find the gradient of any point on any graph, whether it is a quadratic, cubic or any other type of graph. The Formula To find the gradient of a point on a graph, you should use this formula: When y=ax b +c, dy/dx=abx b-1. Dx/dy is the gradient of the point, when you replace the x in abx b-1 with the x-coordinate of the point on the graph. Using the above formula, you can find the equation of the gradient for any graph, such as: Example Question Find the gradient of y=x 2 +4x+7 at (3, 28). y=x 2 +4x+7 dy/dx=2x+4 2*3+4=10 The gradient of the point is 10. When y=x 2 +3x+2, dy/dx=2x+3. When y=x 4 +2x 3-4x 2 +4, dy/dx=4x 3 +6x 2-8x Finding a point with a specific gradient In order to find a point on a graph that has a specified gradient, you merely need to reverse the process find the equation of the gradient, solve for x with dy/dx being equal to zero, then solve the equation of the graph to find y. Example Question Find the point on y=x 2 +4x+7 where the gradient is 10. y=x 2 +4x+7 dy/dx=2x+4 10=2x+4 6=2x 3=x y=3 2 +4*3+7 1

13 Maths Revision Differentiation y=28 The point with gradient 10 is (3, 28). Fractional Equations If you have to find the equation of the gradient of a graph like y=4/x, use this formula to simplify the equation: When y=a/x b, y=ax -b. In the above example of y=4/x, you would first simplify the graph to y=4x -1. Therefore, the equation of the gradient would be dy/dx=-4x -2. You can then reverse the formula if the original form of the equation would be easier to solve (in this case, dy/dx=-4/x 2 ). 2

14 What they are and how to use them.

15 Learning Objectives By the end of this lesson you should: Know what a Second Derivative is Know where to use them Know how to use them

16 What is the second Derivative? The second derivative is what you get when you differentiate the derivative. Not so hard is it?

17 Where do I use them? The second derivative is used when your trying to find if a stationary point is a local maximum/minimum or if its just a point of inflection.

18 How? Simple, differentiate your derivative. As you already know how to differentiate this should be no trouble. A stationary point on a curve occurs when dy/dx = 0. Once you have established where there is a stationary point, the type of stationary point (maximum, minimum or point of inflexion) can be determined using the second derivative. If d2y/dx2 is positive, then it is a minimum point. If d2y/dx2 is negative, then it is a maximum point. If d2y/dx2 is zero, then it could be a max, a min or a point of inflexion. If d2y/dx2=0, you must test the values of either side of the stationary point, as before.

19 By Tariq Willis

20 Roots of equations The definition of a root of an equation is: The value which, is substituted for the unknown quantity in an equation, it this satisfies the equation. For example, x = 0 and x = 5 are roots of the equation x 2-5x = 0.

21 Applications of Roots of an Equation We have already learnt how to find the points of intersection of y=f(x) and the line y=k. We can solve this by solving the equation f(x)=k However frequently we will want to reverse this procedure as we will start with the equation f(x) and we will have to draw the graph y=f(x) and y=k will tell us something about the roots. The x-coordinate of the points of crossing are the roots of the equation. So if we know the shapes of graphs then we can easily find out how many roots there are. There are some illustration on the next slide.

22 y Graphs y y=f(x) y=f(x) y=k x x y A graph with 2 roots y=k y=k Both of the points above are positive, thus there is two roots as we cannot have negative roots. y=f(x) This graph as 1 root. x This graph has no roots.

Learning Target: I can sketch the graphs of rational functions without a calculator. a. Determine the equation(s) of the asymptotes.

Learning Target: I can sketch the graphs of rational functions without a calculator Consider the graph of y= f(x), where f(x) = 3x 3 (x+2) 2 a. Determine the equation(s) of the asymptotes. b. Find the