1. http://www.thescienceforum.com/Calcu...ions-8994t.php

Ok, starting on the chapter 3 review.

Calculate y'

7) y= e<sup>sin 2Θ</sup>

From what I currently know, I would conclude that:

y' = e<sup>sin 2Θ</sup>

since if

a = e<sup>b</sup>

then

a' = e<sup>b</sup>

However, since the book gives the answer as

(2 cos 2Θ)e<sup>sin 2Θ</sup>

then I suspect that there must be a method involved to take the derivative of the exponent of e to get that answer, however I have not yet found it.

2.

3. So y = e<sup>sin 2Θ</sup> is the composition of a few functions. If you want to look at it really carefully, we could define:

f(Θ) = e<sup>Θ</sup>
g(Θ) = sin Θ
h(Θ) = 2Θ

Then our function is y = f(g(h(Θ))). Now how do we take the derivative of composite functions?

[Note: you're right that, if y = e<sup>Θ</sup>, then y' = e<sup>Θ</sup>. The problem in our case is that the exponent isn't just Θ--it's a function of Θ.]

4. well, at least with two functions involved:

(f(g(x)))' = f '(g(x))(g '(x))

I think I am just queezy taking the derivitive of an exponent

But really, I think I might have difficulty now applying the chain rule to 3 functions, because I just know that this work is not right:

y' = (e<sup>sin 2Θ</sup>)(cos 2Θ)

OH! wait I got it!

f(Θ) = e<sup>Θ</sup>, f '(Θ) = e<sup>Θ</sup>
G(Θ) = sin 2Θ
g(Θ) = sin Θ, g '(Θ) = cos Θ
h(Θ) = 2Θ, h '(Θ) = 2
y = e<sup>sin 2Θ</sup>
y'=f '(G(Θ))(G '(Θ))
y'=f '(G(Θ))(g '(h(Θ))(h '(Θ)))
y'=(e<sup>sin 2Θ</sup>)(cos 2Θ)2
y'=(2 cos 2Θ)(e<sup>sin 2Θ</sup>)

5. Great! You just have to apply the chain rule a couple of times, which can be a little tricky (it takes a certain amount of bookkeeping), but you'll get used to it with practice.

6. Now I think I am having problems with implicit differentiation. I don't think I really understand why y<sup>3</sup> = y'(3y<sup>2</sup>). Anyway, here is the review question:

Calculate y'
15) xy<sup>4</sup> + x<sup>2</sup>y = x + 3y

I thought it would be easier to solve if I could get y on one side and simplify as best I could for y. Here is my work:

xy<sup>4</sup> + x<sup>2</sup>y - 3y = x
y ( xy<sup>3</sup> + x<sup>2</sup> - 3 ) = x
xy<sup>3</sup> + x<sup>2</sup> - 3 = x / y
xy<sup>3</sup> - x / y = - x<sup>2</sup> + 3
xy<sup>3</sup> - x / y = 3 - x<sup>2</sup>
x ( y<sup>3</sup> - y<sup>-1</sup> ) = 3 - x<sup>2</sup>
y<sup>3</sup> - y<sup>-1</sup> = x<sup>-1</sup>( 3 - x<sup>2</sup> )

Now I differentiate both sides:

( y ')( 3y<sup>2</sup> ) + ( y ')( y<sup>-2</sup> ) = -2x ( x<sup>-1</sup> ) + ( 3 - x<sup>2</sup> )( -x<sup>-2</sup> )
( y ')( 3y<sup>2</sup> ) + ( y ')( y<sup>-2</sup> ) = -2 - 3x<sup>-2</sup> + 1
( y ')( 3y<sup>2</sup> + 1 / y<sup>2</sup> ) = - 3 / x<sup>2</sup> - 1
( y ')( ( 3y<sup>4</sup> + 1 ) / y<sup>2</sup>) = ( -3 - x<sup>2</sup> ) / x<sup>2</sup>
y '= ( ( -3 - x<sup>2</sup> )( y<sup>2</sup> ) ) / ( ( 3y<sup>4</sup> + 1 )( x<sup>2</sup> ) )

However the answer the book comes up with is:

( 1 - y<sup>4</sup> - 2xy ) / ( 4xy<sup>3</sup> + x<sup>2</sup> - 3 )

7. I'm pretty sure you have to do some manipulation here to make your answer and the book's agree. I haven't done the calculations, but it looks to be that you can use the original equation to make some substitutions in, say, your answer to make it look like the book's. In any case, I cannot find any error in your work.

When we're differentiating implicitly, we have to think of x and y as functions of x. x is a function whose derivative is 1. y is a function whose derivative is y'. So when we take derivatives of something involving y, we'll get a y' by the chain rule. For example, (y^3)' is y'(3y^2) by the chain rule: y is the inner function, and cubing is the outer function, so you take the derivative of the inner (y') times the derivative of the outer (squaring then multiplying by 3) applied to the inner (3y^2). And when you differentiate something like xy^4, you have to use the product rule for the two functions x and y^4. So you get:

(xy^4)' = x'y^4 + x(y^4)' = y^4 + xy'(4y^3) = y^4 + 4xy'y^3

xy^4 is the first term of the original equation. I recognize the quantities on the right as terms in the book's answer (y^4 is on top, 4xy^3 is on bottom), so that suggests to me the book approached the problem by just differentiating the original equation.

Edit: added things in bold to make this post more sensical.

8. I am pretty sure this question will be out of my league with my current understanding, but why is the derivitive of x, a domain value or an independent variable, already known to be 1 while the derivitive of the range value or dependent variable is only known by the symbol y'? Maybe that's not the question I really want to ask, but seeing y' not as an expression but part of an equation is really starting to make me think of the possible implications. What does it mean to say y'(3y<sup>2</sup>)?... y squared, then multipied by 3, then multipied by the change of y. I just get the feeling that there is something behind this that I don't see, but only think is there.

I've been doing my implicit differentiation very long and elaborate to make sure I don't miss anything. I took an hour to differentiate x<sup>2</sup> cos y + sin 2y = xy. I broke it down into 18 functions including the differentiated functions.

I'm sorry if this post seemed to ramble too much.

9. Ok, here is one I am stuck on.

Calculate y'

25) sin(xy) = x<sup>2</sup> - y

my work:

F(x) = sin xy
f(x) = sin x
f '(x) = cos x
g(x) = xy
g '(x) = y + xy'
F(x) = f(g(x))
G(x) = x<sup>2</sup> - y
G '(x) = 2x - y'
F '(x) = f '(g(x))(g '(x))
F '(x) = cos ( y + xy')

sin(xy) = x<sup>2</sup> - y
F(x) = G(x)
F '(x) = G '(x)
cos ( y + xy' ) = 2x - y'

then maybe:
( cos y )( cos xy' ) - ( sin y )( sin xy' ) = 2x - y'
( cos y )( cos xy' ) - ( sin y )( sin xy' ) + y' = 2x

but I don't see how to go any farther, unless there is a way to seperate something such as cos xy.

10. Regarding your less recent post, you could think of x and y as "functions of x". x is just the function f(x) = x. Since f'(x) = 1, we have x' = 1. y is some function which is given implicitly by your equation; you can think of it as some function y(x), and we're just suppressing the dependence on x by writing y. Since we don't know anything about y except that it's given implicitly by some equation in x and y, the best we can say about y when differentiating is that it is some function y' = y'(x).

So thinking like this, let's look at x<sup>2</sup>cos(y) + sin(2y) = xy. It's good to be able to do this really carefully, but it's also important to become so comfortable with the rules of differentiation that you can do them in your sleep. Let's look at the terms x<sup>2</sup>cos(y), sin(2y), and xy independently. The trickiest is probably the first...

(x<sup>2</sup>cos(y))'
= (x<sup>2</sup>)'cos(y) + x<sup>2</sup>(cos(y))' [product rule]
= 2x cos(y) + x<sup>2</sup>(cos(y))' [power rule]
= 2x cos(y) + x<sup>2</sup>cos'(y)y' [chain rule]
= 2x cos(y) - x<sup>2</sup>sin(y)y' [cos' = -sin]

(sin(2y))'
= sin'(2y)(2y)' [chain rule]
= cos(2y)(2y)' [sin' = cos]
= 2 cos(2y) y' [constants pull out]

(xy)'
= x'y + xy' [product rule]
= 1y + xy' [x' = 1]
= y + xy'

I'm sure you can do the rest from here.

As for your most recent post, stuff looks good until you plug things in to calculate F'(x). You (correctly) have:

-F'(x) = f'(g(x))(g'(x))
-f'(x) = cos(x)
-g(x) = xy
-g'(x) = y+xy'

But somehow you conclude that:

-F'(x) = cos(y+xy')

Check your work more carefully. You should never end up with a y' inside of a function; to see this, just apply all of the differentiation rules for arbitrary functions with y as one of the functions. All terms should be either functions of x and y alone or (functions of x and y) times y'. Then you can always put the former on one side, the latter on the other, and solve for y'.

Implicit differentiation is not the most important skill to learn. Most of what you'll be dealing with will be equations of the form y = f(x). Don't give yourself a stroke trying to figure all of these out... focus more on getting the rules of derivatives (especially the chain rule) down.

11. I think what I did was misread g(x) and g '(x). Where I should have had f '(g(x)), I had f '(g '(x)).

Calculate y'
28 ) y = (cos x)<sup>x</sup>

Now my usual method of breaking down the equation into simplier functions and the derivitives of those functions gets thrown a curveball. Here is my work:

y = F(G(x))
G(x) = cos x

Now I very well can't say F(x) = x<sup>x</sup> as I would like to, because that would make y = (cos x)<sup>(cos x)</sup> which is not true. What I did decide to do was impilment a third variable, z.

F(z) = z<sup>x</sup>

With this I believe that y = F(G(x)), but I am not completely sure how to get F '(z), so I guessed that perhaps

F '(z) = xz<sup>(x-1)</sup>

Although this seems to account for the derivitive of the x value, I suspect that z should also have more of a part in the derivitive. Anyway, continuing on:

y' = F '(G(x))(G '(x))
y' = (x cos x)<sup>(x-1)</sup>(-1 sin x)

12. This is a tricky one which requires a little bit of ingenuity. Your solution runs into problems because, at one point, you're considering x as a variable (in G(x)), and at another, as a constant (in F(z)).

The book probably goes over a few examples that look like this. But the right idea is to take logs of both sides:

ln y = ln(cos x)^x = x ln(cos x)

Then you can use implicit differentiation.

(In this case, you can get down to an explicit formula for y' in terms of x, as your original equation gives you y in terms of x and implicit differentiation gives you y' in terms of x and y.)

13. Ok. Thanks, so:

y = (cos x)<sup>x</sup>
ln y = ln ((cos x)<sup>x</sup>)

and since ln b = log<sub>e</sub>b, and log<sub>a</sub>b<sup>r</sup> = r log<sub>a</sub>b, then if z = b<sup>r</sup> then ln z = r ln b and:

ln y = x ln cos x

Then I can brake it down as I like:

F(x) = ln y
G(x) = x ln cos x
f(x) = ln x
g(x) = y
h(x) = x
j(x) = ln cos x
k(x) = cos x
f '(x) = 1/x
g '(x) = y'
h '(x) = 1
k '(x) = -sin x
j '(x) = f '(k(x))k '(x) = -sin x / cos x

G(x) = h(x)j(x)
G '(x) = h '(x)j(x) + h(x)j '(x)
G '(x) = ln cos x - x sin x / cos x
F(x) = f(g(x))
F '(x) = f '(g(x))g '(x)
F '(x) = y' / y
F '(x) = G '(x)
y' / y = ln cos x - x sin x / cos x
y' = y ( ln cos x - x tan x )

By the way, how do you verbally say y' in a short verbally efficient way? Do you just usually say "the derivitive of y, blah blah blah, times the derivitive of y squared, ect?"

14. Here is another one I am not getting right.

Calculate y'
31) y = x tan<sup>-1</sup> (4x)

As far as I know, it wouldn't hurt to change tan<sup>-1</sup> into cot, but I have done it both ways.

y = x cot 4x

y '= cot 4x + x (cot 4x) '
y ' = cot 4x + x (4(-csc 4x)<sup>2</sup>)
y ' = cot 4x - 4x csc<sup>2</sup> 4x

or I could say:

y ' = ( cos 4x sin 4x - 4x ) / sin<sup>2</sup> 4x

The answer in the book is:
4x / (1+ 16x<sup>2</sup>) + tan<sup>-1</sup> 4x

The second part of the answer agrees with my answer, but according to the answer in the book x(cot 4x)' does not equal - 4x csc<sup>2</sup> 4x

15. You can pronounce y' as "y prime". You can also call y'' "y double prime" and extend this by analogy to "y triple prime" and the like.

Now there is an important convention you need to learn when dealing with trigonometric functions. It looks like you know part of the convention: whenever we write something like:

tan<sup>n</sup> x

where n > 0, we mean:

(tan x)<sup>n</sup>

But when we write:

tan<sup>-1</sup> x

we really mean the inverse function to tangent (in the sense of that old discussion involving you, Guitarist, and myself).

So for your problem, we have a couple of methods of approach. The first is to make tan<sup>-1</sup> disappear, and this will involve applying tangent to the equation somehow. In fact, I would suggest:

y = x tan<sup>-1</sup>(4x)
y/x = tan<sup>-1</sup>(4x)
tan(y/x) = 4x

Then you can apply implicit differentiation. This looks a little tricky to me, though. It could be easier if you first figured out what the derivative of tan<sup>-1</sup> x is. To do this, you would use implicit differentiation on the equation:

tan y = x

and you'll probably have to use sec<sup>2</sup> = 1 + tan<sup>2</sup> to get this into a nice form. (Eventually you be expected to know this derivative, so it's worth going through the steps anyway.) Knowing this, you can just differentiate the original function as normal.

16. not specifically related to this problem, should you not define 1/tan by tan<sup>-1</sup>? That would mean that when you take the derivitive of something like Sqrt(tan x), you would write it 1/(2Sqrt(tan x)) instead of (1/2)Sqrt(tan x)^(-1/2). It seems confusing if both should be written the same way. Then what happens when you get something like tan<sup>-2</sup> x, or -3 or -4? Do you just keep swapping x and y if tan x = y? Why not just write it as arctan to avoid confusion?

17. I agree that the exponent-attached-to-trig-function convention is inconsistent. As you suggest, many mathematicians prefer using arctan over tan<sup>-1</sup> for this reason. For some reason, Stewart refuses to address this problem, so you'll just have to remind yourself that tan<sup>-1</sup> means arctan whenever you read his book. Also, mathematicians do have a way of writing 1/tan. It's called cotangent!

Now let me exposit upon the convention of using f<sup>-1</sup> to mean 1/f versus the convention of using f<sup>-1</sup> to mean f<sup>-1</sup>⋅f(x) = x (where the little dot is supposed to be function composition). The meaning is usually clear from context. For example, suppose I am dealing with completely abstract sets X and Y and some function f: X -> Y. Since Y isn't supplied with any sort of arithmetic operations, we'd assume that f<sup>-1</sup> means the inverse function if it exists (or, if the function is not invertible, the "preimage" function which sends y to the set of x with f(x) = y).

If suddenly Y is identified to be the real numbers R, we're probably going to assume that f<sup>-1</sup> means 1/f. (This is inconsistent with the trig function notation, but bear with me for a second.) The reason for this is that the functions from X to R are a nice "algebra" of functions--you can add and multiply pointwise to define an addition and multiplication on the functions. When dealing with this sort of object, we're probably more interested in these properties than in taking preimages.

Now if X and Y are both R, and we're restricting ourselves to functions which are "bijective", i.e. invertible and so that the whole range is equal to R, then there's a good chance we're interested in f and g as somehow "preserving the structure of R"--maybe in a geometric or arithmetic sense. In this case, we'd again want to let f<sup>-1</sup> to mean the inverse function, because now we can forms a sort of algebraic object called a "group" out of these functions--we can compose any two and get a third function with the same properties, and we can always invert them, and the composition of a function and its invert, by definition, gives us the identity function (i.e., Id(x) = x).

Okay, these last three paragraphs are a lot to digest, I imagine, and probably not helpful to what you're doing right now, but if you plan to continue into more abstract mathematics after getting through calculus, beginning to think about such things may not be a bad idea.

18. Yea, I'll look up your last post once in a while. I was starting to think about what is beyond calculus, but I'll save those questions for later.

Here is one that seems like it should be very simple:

47) Calculate y '

y = cosh<sup>-1</sup> (sinh x)

My work:

Getting rid of the inverse:
sinh x = cosh y

Implicit differentiation:
cosh x = y '(sinh y)
y '= cosh x / sinh y

( cosh x ) / Sqrt( sinh<sup>2</sup> x - 1)

19. Ok. Back to:

31) y = x tan<sup>-1</sup> 4x

My new work:

f(x) = x
g(x) = tan<sup>-1</sup> x
h(x) = 4x
f '(x) = 1
h '(x) = 4

to find g '(x):

z = g(x)
z '= g '(x)
tan<sup>-1</sup> x = z
tan z = x
z 'sec<sup>2</sup> z = 1
z '= 1/sec<sup>2</sup> z
z '= 1/(tan<sup>2</sup> z +1)
z '= 1/(tan<sup>2</sup> (tan<sup>-1</sup> x) + 1)
z '=1/(tan x + 1)
g '(x) = 1/(tan x +1)

y = f(x)g(h(x))
y '= f '(x)g(h(x)) + f(x)g '(h(x))h'(x)
y '= ( 1 )( tan<sup>-1</sup> 4x ) + ( x )( 1 / ( ( tan 4x) + 1 )( 4 )
y '= tan<sup>-1</sup> 4x + 4x / ( ( tan 4x ) +1 )

The book still disagrees with its answer:

4x / (1+ 16x<sup>2</sup>) + tan<sup>-1</sup> 4x

20. 47. So you come up with:

y' = cosh x/sinh y

and the book has:

y' = cosh x/(sinh<sup>2</sup> x - 1)<sup>1/2</sup>

How did the book get this? They got rid of y by using the original equation. Let's use it in the form:

cosh y = sinh x

The denominator has sinh y in it. We need to relate cosh y to sinh y somehow. We do this by the "hyperbolic analog of the Pythagorean theorem":

cosh<sup>2</sup> - sinh<sup>2</sup> = 1

So we have:

sinh y = (cosh<sup>2</sup> y - 1)<sup>1/2</sup> = (sinh<sup>2</sup> x - 1)<sup>1/2</sup>

This is precisely the denominator the book gives. (This also looks a lot like one of the steps you used in number 31, when you replace secants with tangents.)

In general, if you are given a function y = f(x) which requires implicit differentiation to find y', Stewart will use the original equation to eliminate the y variable from the formula for y'.

31. Your work is perfect with one exception: you claim that:

tan<sup>2</sup> (tan<sup>-1</sup> x) = tan x

Remember that tan<sup>2</sup> x = (tan x)<sup>2</sup> while tan<sup>-1</sup> x = arctan x. Thus the equality should be...

tan<sup>2</sup> (tan<sup>-1</sup> x) = (tan (arctan x))<sup>2</sup> = x<sup>2</sup>

Normally I'd just say, "And then you can go from there," but I'd like to point out that this is a very important formula you deduce here:

arctan'(x) = 1/(1+x<sup>2</sup>)

This will become much more important when we get to integration. Differentiating algebraic functions (i.e., functions involving only arithmetic operations and root extractions) always yields an algebraic function, but sometimes "transcendental" functions (e.g., arctan) have algebraic derivatives. Integration is in some sense the opposite of differentiation, and so this means that the result of integrating an algebraic function is not always algebraic. So, for example, if you want to integrate rational functions (i.e., ratios of polynomials), you often need to know the integral of 1/x and 1/(1+x<sup>2</sup>), which are ln x and arctan x.

21. I've been studying for a couple days, and this is where I am at right now:

I havn't quite gone the direction you last specified I think just yet on problem 47. I decided first to review a bit on how I had previously worked out inverse trig functions, and decided to use the old triangle diagram to help myself better understand the processes described within the equation for problem 47. Originally this lead me to the following work:

y' = cosh x / sinh y
y' = cosh x / ( sinh ( cosh<sup>-1</sup> ( sinh x ) ) )

At this point I created a right triangle with an angle x, and an angle ( x - π/2 ). I assigned a third variable to the opposite side of angle x which I named as z, and the hyponenuse as 1, thus making the sin of x = z. So:

sin x = z

So now we have:

y' = cosh x / ( sinh ( cosh<sup>-1</sup> z ) )

By the triangle diagram we can deduce that cosh<sup>-1</sup> z = ( x - π/2 ), thus:

y' = cosh x / sinh (x - π/2)

However since this answer seems much different from the book answer, it must be wrong. Why was it wrong I asked myself, and my answer was that hyperbolic functions, while similar to circular functions, might act very different. It is these differences and similarities that I am trying to understand at the moment.

22. The problem with your reasoning is you're using properties of the circular trig functions to manipulate the hyperbolic ones. There is a geometric way of working with the hyperbolic trig functions, but it requires different triangles. This isn't a standard part of the curriculum nowadays (and hasn't been for hundreds of years, I imagine), and do I know much about it myself.

(One minor thing... if we were indeed working with circular trig functions, your triangle diagram should have angles of x, π/2 - x, and π/2 so as to have the angles add up to π.)

Everything you are expected to do with the hyperbolic functions in Stewart is essentially algebraic, so you don't need to know the geometric aspect. He defines them in terms of exponentials, and so all of their properties can be studied via algebra and the properties of the exponential function. For example, the "Pythagorean theorem" I stated is a simple consequence of the definitions of sinh and cosh.

23. I'll work it out your way soon, but in an effort to try to understand what it means to say sinh x, or cosh x, and such, I ended up going a different way I think. Because I based my understanding of sin, cos, tan mainly on geometric principles, it seemed more difficult to me to understand what sinh, cosh, and tanh actually mean since my introduction to the concept was based on algebraic principles. However going off of what I have recently learned about the definitions of the hyperbolic functions, I have come up with the following work:

47)

y' = cosh x / sinh y
y' = cosh x / ( sinh ( cosh<sup>-1</sup> ( sinh x ) ) )

by definition:
sinh x = ( e<sup>x</sup> - e<sup>-x</sup> ) / 2

then:
cosh<sup>-1</sup> ( (e<sup>x</sup> - e<sup>-x</sup>) / 2 ) = z
cosh z = ( e<sup>x</sup> - e<sup>-x</sup>) / 2

by definition:
cosh z = ( e<sup>z</sup> + e<sup>-z</sup>) / 2

then:
( e<sup>z</sup> + e<sup>-z</sup>) / 2 = ( e<sup>x</sup> - e<sup>-x</sup>) / 2
e<sup>z</sup> + e<sup>-z</sup> = e<sup>x</sup> - e<sup>-x</sup>
(e + e<sup>-1</sup>)<sup>z</sup> = (e - e<sup>-1</sup>)<sup>x</sup>
log<sub>( e + e^-1 )</sub>( e - e<sup>-1</sup> )<sup>x</sup> = z
x log<sub>( e + e^-1 )</sub>( e - e<sup>-1</sup> ) = z

cosh<sup>-1</sup> ( ( e<sup>x</sup> - e<sup>-x</sup> ) / 2 ) = x log<sub>( e + e^-1 )</sub>( e - e<sup>-1</sup> )

and finally:
y' = cosh x / ( sinh ( x log<sub>( e + e^-1 )</sub>( e - e<sup>-1</sup> ) )

Yes, previously I did make a mistake on that right triangle angle by making the angle opposite to x equal to x - π when it should have been π - x.

24. In general we don't have the equality:

(a+b)<sup>c</sup> = a<sup>c</sup> + b<sup>c</sup>

so a couple of your steps below don't make sense. I tried to work it out following a line of reasoning similar to yours, but I still needed to appeal to the formula cosh<sup>2</sup> - sinh<sup>2</sup> = 1 at some point.

My honest advice is don't stress over hyperbolic functions. They're a drop in the bucket that is Stewart's Calculus. They're certainly worth learning about, and you should pursue this topic on the side, but don't let them prevent you from progressing with the calculus.

25. On page 250, Stewart defines the hyperbolic functions sinh and cosh as:

sinh x = (e<sup>x</sup> - e<sup>-x</sup>) / 2
cosh x = (e<sup>x</sup> + e<sup>-x</sup>) / 2

The definitions of the other four hyperbolic functions are defined in terms of sinh and cosh. I used these definitions to get used to what sinh and cosh actually mean, so when I had sinh x or cosh x I converted this into the terms of e that were stated in the definition.

Using your suggestions, I have worked it out using Hyperbolic Identities. This time my goal was to, instead of finding y in terms of x, find sinh y in terms of x. Here is my work:

47) y = cosh<sup>-1</sup> (sinh x)
sinh x = cosh y

Using identities we know that:
sinh x = (cosh<sup>2</sup> x - 1)<sup>1/2</sup>
cosh y = (sinh<sup>2</sup> y + 1)<sup>1/2</sup>

so if sinh x = cosh y, then
sinh x = (sinh<sup>2</sup> y + 1)<sup>1/2</sup>
sinh<sup>2</sup> x = sinh<sup>2</sup> y + 1
sinh<sup>2</sup> x -1 = sinh<sup>2</sup> y
( sinh<sup>2</sup> x -1 )<sup>1/2</sup> = sinh y

so if y' = cosh x / sinh y, then

y' = cosh x / ( sinh<sup>2</sup> x -1 )<sup>1/2</sup>

51) Find y'' if x<sup>6</sup> + y<sup>6</sup> = 1

Soving implicitly:

6x<sup>5</sup> + 6y<sup>5</sup>y' = 0
x<sup>5</sup> + y<sup>5</sup>y' = 0
or:
y' = -x<sup>5</sup> / y<sup>5</sup>

Is this next step correct?
5x<sup>4</sup> + 5y<sup>4</sup>y' + y<sup>5</sup>y'' = 0
y'' = ( -5x<sup>4</sup> - 5y<sup>4</sup>y' ) / y<sup>5</sup>

To avoid y' in the answer perhaps I could have first solve for y:
y = ( 1 - x<sup>6</sup>)<sup>1/6</sup>

Differentiating:
y' = -x<sup>5</sup>( 1 - x<sup>6</sup> )<sup>-7/6</sup>
y'' = ( -7x<sup>10</sup>)( 1 - x<sup>6</sup> )<sup>-13/6</sup> - 5x<sup>4</sup>(1 - x<sup>6</sup>)<sup>-7/6</sup>

5x<sup>4</sup> / y<sup>11</sup>

So far from what I have learned I think the only way to get y in the answer of a differentiation is if it was differentiated implicitly, which must mean that my implicit differentiation here is incorrect. Perhaps Stewart subsituted y', so going back to my implicit solving:

6x<sup>5</sup> + 6y<sup>5</sup>y' = 0
y' = -x<sup>5</sup> / y<sup>5</sup>

so if this is right:
5x<sup>4</sup> + 5y<sup>4</sup>y' + y<sup>5</sup>y'' = 0

then after substitution for y' and simplifying:
y'' = ( 5x<sup>5</sup> - 5x<sup>4</sup>y ) / y<sup>6</sup>
y'' = 5x<sup>4</sup>( x - y ) / y<sup>6</sup>

which aparently is also incorrect.

26. I just realized I never suggested to you that I, too, was working from the "exponential definitions" of the hyperbolic functions. The identity I was using can be derived from these definitions. Sorry for the confusion!

For 51, there's a small problem with the book. The answer should be:

y'' = -5x<sup>4</sup>/y<sup>11</sup>

(I forget if the book has covered concavity yet, but if it has, you can tell that my answer at least gives the correct sign versus the book's.) Next, there is a minor error in your first calculation. You start correctly with:

x<sup>5</sup> + y<sup>5</sup>y' = 0

and you take the derivative of this equation. When you take the derivative of y<sup>5</sup>y', you should get:

(y<sup>5</sup>y')' = (y<sup>5</sup>)'y' + y<sup>5</sup>y'' = (5y<sup>4</sup> y')y' + y<sup>5</sup>y'' = 5y<sup>4</sup>(y')<sup>2</sup> + y<sup>5</sup>y''

If you use this, you'll get an expression which you can manipulate to look like the book's.

27. Damn, I worked so hard on that problem, and tried to get it so accurate, but I forgot that small little y' near the end. Anyway, I've been going through the problems much faster now. It is seeming like I almost completely know the Chapter 3 material. In this problem:
-----------------------------------------------
66b) By differentiating the addition formula

sin(x+a) = sin x cos a + cos x sin a

obtain the addition formula for the cosine function.
------------------------------------------------
should I consider a as a constant, dependent variable, or independent variable?

28. I want to make sure I have this one right too:

Find f ' in terms of g '.
74) f(x) = e<sup>g(x)</sup>

My work:

f(x) = y
ln y = ln g(x)

Differentiating:
y'/y = g'(x) / g(x)
y' = ( y ) g'(x) / g(x)
y' = f(x)g'(x) / g(x)

If we wanted to substitute f(x) then:
f '(x) = e<sup>g(x)</sup>g'(x) / g(x)

29. I think I am going to move on to the next chapter now. I think I've got all this Chapter 3 stuff down. It seems all of the review questions that are left are more or less just word problems to make sure that you understand how the material might apply to real life problems, but I don't think it is necisary, at least for me. I understand that a derivitive is a rate of change, and just that simple statement should sum up the purpose of the rest of the 20 or so review questions. Unless you think there is anything more I should know about this Chapter 3, I think I'll move on to Chapter 4: Applications of Differentiation.

Oh, and concavity is covered in 4.3.

30. For the trig identity question, you should think of a as being constant--we aim to get a formula for cos(x+a), and cos(x+a) is just the derivative of sin(x+a) if we hold a constant.

For the next question, you let:

f(x) = y

and then you say:

ln y = ln g(x)

But if:

f(x) = e<sup>g(x)</sup>

then shouldn't:

ln y = g(x) ?

In any case, I think you should be ready to move on to chapter 4. Just be sure to look at the next couple of chapters not only as new material but also as an opportunity to hone your differentiation skills.

 Bookmarks
Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement