CHAPTER 1. DIFFERENTIATION 18. As x 1, f(x). At last! We are now in a position to sketch the curve; see Figure 1.4.

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1 CHAPTER. DIFFERENTIATION 8 and similarly for x, As x +, fx), As x, fx). At last! We are now in a position to sketch the curve; see Figure.4. Figure.4: A sketch of the function y = fx) =/x ). Observe the stationary point at x =0; the fact that this is a maximum has been deduced with the help of the vertical asymptotes. Example.9. Sketch the graph of Again, we follow the recipe... y = x x) 4 x,.3) ) Note that y = x x) x) + x), therefore there are vertical asymptotes at x = ±. Also, are only interested in real y, thuswerequirey > 0. Hence it follows that y is defined only when x x) 4 x > 0. The RHS of.3) may change sign at x =0,, and possibly at the position of the vertical asymptotes! Consider the following diagram of the sign of y : Therefore the graph of y is undefined for ) y is neither odd nor even, but observe x<0 and <x. y = ± x x) 4 x and the ± sign indicated that the graph should be symmetric about the horizontal x axis.

2 CHAPTER. DIFFERENTIATION 9 Figure.5: You can make a sign diagram for y = x x) x)+x),too! Becausey is nonnegative for any real value of y, the function is undefined wherever we find that y < 0 these are indicated by a minus sign in the diagram). 3) y =0whenx =0,. 4) x =0 y = 0 but we already know that!). 5) dy dx is stationary when d dx y )is,since d dx y )=y dy dx. d dx y )= 4 x ) x) x x ) x) 4 x ) =0. For this to be zero the numerator must be zero. Therefore simplifying the numerator leads to x 8x +4=0 x =4± , 7.5). Rather than calculating the second derivative which would be quite tedious), we can deduce the nature of these turning points from the information regarding the behaviour near the horizontal asymptotes. 6i) To figure out the behaviour of the behaviour as x ±,write and use the geometric series y = x 4 x.4) z =+z + z +..., for z <, so Equation.4) can be approximated as for large x ) y )+ 4x ) x +... x,.5) which is valid for x.thus As x, y from below) As x, y + from above) In addition, there are there are mirror images see Step ) of this horizontal asymptote, i.e. at y =.

3 CHAPTER. DIFFERENTIATION 0 Figure.6: Plots of the upper branch of fx) for x< and 3 <x<9 respectively. 6ii) To get the behaviour near the vertical asymptotes it is simplest in this case) to find where the curve cuts its horizontal asymptote, i.e. set y = : 4 x = x x x =4 Hence we can sketch two parts of the upper half of the graph, see Figure.6. And let s not forget to plot the rest of the graph! Figure.7: The complete sketch for the implicit) function y = x x) 4 x..5.3 Equations of Tangent and Normal Example.30. Find equations of the tangent and normal to y = x at x =. First find dy dy dx, recalling that dx slope of the tangent. dy dx =x, dy =. dx x= Also, at x =wehavey =. Therefore using y y = mx x )

4 CHAPTER. DIFFERENTIATION where x =,y = and m =, the line through, ) with slope has equation y =x. The normal is perpendicular to the tangent. Therefore Slope of Normal = Slope of Tangent =. The normal is the line through, ) with slope = /. Therefore using y y = mx x ) with x =,y = and m = / yields the equation for the normal as y = x + 3. Example.3. Find equations of the tangent and normal to the curve given by For this we use parametric differentiation y = t, x = t 3 + at t =. dy dy dx = dt dx dt = t 3t = 3 at t =. Also at t =, x, y) =, ). The tangent is the line through, ) with slope 3,i.e. y = 3 x ), y = 3 x 3. The normal has slope 3, and thus its equation is y = 3 x ), y = 3 x +4.

5 Chapter Hyperbolic functions. Definitions of hyperbolic functions In the first chapter, we got a few glimpses of hyperbolic functions, so now you re probably itching to find out just what they are. Well, that s what this chapter is for! First things first, here are the definitions: sinh x = ex e x cosh x = ex + e x tanh x = ex e x sinh x e x = + e x cosh x. The three functions are pronounced shine x, cosh x and tansh x respectively. Recall that as x, e x and e x 0. If y = cosh x = ex +e x, Also note that cosh 0) =. y = cosh x) = e x + e x) = e x + e x = cosh x. Therefore the curve is symmetrical about the y axis, i.e. is an even function. And as x, y ex +0 = ex. If y =sinhx = ex e x, Also, sinh 0) = 0. y =sinh x) = e x e x) = e x e x = sinh x,

6 CHAPTER. HYPERBOLIC FUNCTIONS 3 therefore the curve is anti-symmetrical about the y axis, i.e. is an odd function. And as x, y ex 0 as x, y 0 e x = ex +, = e x. 3 For we see that y = tanh x = ex e x sinh x e x = + e x cosh x, tanh 0) = 0 =0. Also, if we consider the limits x ± : Finally, note that so tanh x is an odd function. as x, y ex 0 e x +0, as x, y 0 e x. 0+e x sinh x) tanh x) = cosh x) = sinh x cosh x = tanh x, Figure.: Plots of the three main hyperbolic functions. The blue curve is sinh x, thered curve is cosh x, and the green curve is tanh x.

7 CHAPTER. HYPERBOLIC FUNCTIONS 4. Inverse hyperbolic functions The hyperbolic functions do come with inverse functions. Suppose that Then by definition, y =sinh x, x =sinhy. Multiplying by e y gives or x = which is a quadratic equation in e y. thus Now e y > 0 for all y, but because e y e y) e y e y =x e y xe y =0, e y ) xe y ) =0, e y = x ± 4x +4 = x ± x +,. e y = x + x +, or e y = x x +. x x +< 0, x +>x x +> x = x. So the second option negative choice) is impossible! Hence we are left with e y = x + x +, or y =sinh x =ln x + ) x +. Suppose that Then by definition of cosh, As before, multiply by e y to get y = cosh x, x = cosh y, so x ). e y + e y) = x e y + e y =x e y + xe y =0 or e y ) xe y )+=0.

8 CHAPTER. HYPERBOLIC FUNCTIONS 5 which is a quadratic equation in e y again!) e y = x ± 4x 4 = x ± x, and this is real since x anyway. Therefore Now e y > 0 for all y, and e y = x + x, or e y = x x. x ± x > 0 are both possibilities so we can t rule any option out!) Observe that x + x = x + x x x x x = x x x x ) = x x. Thus So or i.e. e y = x + x or e y = ) y =ln x + x y =ln x + ) x, x + x. = ln x + ) x, y = ± ln x + ) x. Figure.: Plot of cosh x. Note that for a given value of y there are two possibilities for x

9 CHAPTER. HYPERBOLIC FUNCTIONS 6.3 Hyperbolic identities Just like the trigonometric functions, the hyperbolic ones come with all sorts of weird and wonderful identities. You will see many of them in this section. Now is a good time to introduce three more hyperbolic functions. They are... coth x ) tanh x c.f. cot x tan x sech x ) cosh x c.f. sec x cos x cosech x ) sinh x c.f. cosec x sin x.).).3)... and they are pronounced coth, shec and coshec respectively. From the definitions of sinh x and cosh x, and similarly cosh x +sinhx ex + e x + ex e x e x, therefore i.e. cosh x sinh x ex + e x which is analogous to cos x +sin x. ex e x e x, cosh x +sinhx) cosh x sinh x) e x e x cosh x sinh x, Now divide the above result by sinh x to yield cosh x sinh x sinh x, cosech x coth x, which is analogous to cosec x cot x + ). Recall that Squaring both of these yields and then doing.4) minus.5) yields cosh x +sinhx e x cosh x sinh x e x. cosh x +sinhx cosh x +sinh x e x.4) cosh x sinhx cosh x +sinh x e x.5) 4sinhx cosh x e x e x sinhxcosh x ex e x,

10 CHAPTER. HYPERBOLIC FUNCTIONS 7 Hyperbolic Trigonometric coth x / tanh x cot x / tan x sech x / cosh x sec x / cos x cosech x / sinh x sec x / sin x cosh x sinh x cos x +sin x sech x tanh x sec x + tan x cosech x coth x cosec x cot x + sinh x sinhxcosh x sin x sinxcos x cosh x cosh x +sinh x cos x cos x sin x cosh x +sinh x cos x sin x cosh x cosh x cos x cos x Table.: Lots of hyperbolic identities, along with with their trigonometric counterparts. i.e. which is analogous to sin x cos x sin x. sinhx cosh x sinh x, But for now, let s just admire the Table.. Notice that the hyperbolic identities are very similar to the trigonometric counterparts, but with some different signs! This is called Osborne s rule, which tells you to flip the sign whenever we have a product of sinhs; this includes cosech x, tanh x and coth x as well as sinh x! Otherwise the hyperbolic identities are essentially the same as their trigonometric versions. You will get to derive one of these identities as part of your homework!