8: Differentiating some Trig Functions most slides © Christine Crisp

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Presentation transcript:

8: Differentiating some Trig Functions most slides © Christine Crisp

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

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.

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

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 .

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

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

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

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

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

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.

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

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.

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!

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

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

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

2. 3. Let

4. Let

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

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

Summary

Inverse trig functions  

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

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

  rewrite this Answer  

      Answer

    Answer f’(x) = 10 sec²5x tan5x

  rewrite this as sin (x)0.5 Answer is here