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8: Differentiating some Trig Functions most slides © Christine Crisp

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1 8: Differentiating some Trig Functions most slides © Christine Crisp

2 A reminder of the rules for differentiation developed so far!
( I call these functions the simple ones. ) The chain rule ( for functions of a function ): where

3 The trig functions are quite different in shape from either of the simple functions we’ve met so far, so the gradient functions won’t follow the same rules. We’ll start with and use degrees We only need the 1st quadrant as symmetry will then give us the rest of the gradient function.

4 x x x x The function x x x x The gradient function

5 The gradient drops more slowly between and . . .
x x The function x x The gradient function x x x The gradient drops more slowly between and x than between and

6 x x The function x x The gradient function x x x x

7 x x The function x x The gradient function x x x x

8 The gradient function looks like BUT we need a scale on
The function The gradient function looks like BUT we need a scale on the axis. x We can estimate the gradient at x = 0 by using the tangent. So, the gradient is

9 The gradient function isn’t since not .
The function x The gradient function isn’t since not So,

10 However, if we use radians:
x It can be shown that this length . . . is exactly 1. x

11 From now on we will assume that x is in radians unless we are told otherwise.
We have, Exercise Using radians sketch for Underneath the sketch, sketch the gradient function. Suggest an equation for the gradient graph.

12 Solution: This is a reflection of in the x-axis so its equation is

13 SUMMARY If x is in radians, N.B. We have not proved these results; just shown they look correct. We need a bit more theory before we can differentiate the trig function This is done in a later presentation.

14 Compound Trig Functions
We can use the chain rule to differentiate trig functions of a function. e.g. 1 Find the gradient of at the point where N.B. We don’t need to put brackets round 3x. Solution: (a) First find the gradient function: Let N.B. Radians!

15 e.g. 2 Differentiate What would you let u equal in this example? Solution: If we write as we can easily see that the inner function is So, let

16 e.g. 3 Differentiate Be careful! This isn’t the same as Solution: Let

17 Exercise Differentiate the following with respect to x: 1. 2. 3. 4. Solutions: 1. Let

18 2. 3. Let

19 4. Let

20 We can now differentiate the trig function
by writing x y tan =

21 So, This answer can be simplified: is defined as Also, So,

22 Summary

23 Inverse trig functions

24 y = sin (4x + 2) let y = sin u and u = 4x + 2 Answer

25 y = sin x tan 2x use the product rule Answer

26 rewrite this Answer

27 Answer

28 Answer f’(x) = 10 sec²5x tan5x

29 rewrite this as sin (x)0.5 Answer is here

30


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