MHB Extremely Difficult Extremum Problem.

[Amad27]

Here is goes,

Find the minimum value of $y = \left| sin(x) + cos(x) + tan(x) + sec(x) + csc(x) + cot(x) \right|$

I don't even know where to begin.

 

[MarkFL]

Try letting:

$$u=\sin(x)+\cos(x)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$

And writing $$y=f(u)$$. What do you find?

 

[Amad27]

Try letting:

$$u=\sin(x)+\cos(x)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$

And writing $$y=f(u)$$. What do you find?

Ok.

$ y = sin(x) + cos(x) + \frac{sin(x)}{cos(x)} + \frac{1}{sin(x)} + \frac{1}{cos(x)} + \frac{cos(x)}{sin(x)}$

$y = \frac{sin^2(x)cos(x) + cos^2(x)sin(x) + sin^2(x) + cos(x) + sin(x) + cos^2(x)}{sin(x)cos(x)}$

$y = \frac{sin^2(x)cos(x) + cos^2(x)sin(x) + 1 + cos(x) + sin(x)}{sin(x)cos(x)}$

$y = \frac{cos(x)(sin^2(x) + cos(x)sin(x)) + 1 + u}{sin(x)cos(x)}$

We notice

$u^2 = (sin(x) + cos(x))^2 = sin^2(x) + 2sin(x)cos(x) + cos^2(x) = 1 + 2sin(x)cos(x)$

Thus we get later,

$u^2 - 1 = 2sin(x)cos(x) \implies sin(x)cos(x) = \frac{u^2 - 1}{2}$

We are ready for formulate the above $y$ now; to get the following,

$y = \frac{cos(x)(sin^2(x) + \frac{u^2 - 1}{2}) + 1 + u}{\frac{u^2 - 1}{2}}$

But this does get us stuck -_- stuck at cos(sin^2(x))...

 

[mathbalarka]

You made all the substitutions correctly, although you overcomplicated the calculations :

$$y = \sin(x) + \cos(x) + \frac1{\sin(x)} + \frac1{\cos(x)} + \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} $$
$$ = \sin(x) + \cos(x) + \frac{\sin(x) + \cos(x)}{\sin(x)\cos(x)} + \frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)} $$
$$= u + \frac{u}{(u^2-1)/2} + \frac{1}{(u^2-1)/2}$$

 

[anemone]

Hi Olok,:)

You might want to read this thread:

http://mathhelpboards.com/challenge-questions-puzzles-28/minimize-trigonometric-expression-4330.html

 

[Amad27]

Hi Olok,:)

You might want to read this thread:

http://mathhelpboards.com/challenge-questions-puzzles-28/minimize-trigonometric-expression-4330.html

Hello @anemone,

I looked at the thread and I looked a page 2, where you posted on February 2014, for the x = a - 135My question for you,

How in the world did you figure out you should let $x = 1 - 135$

I mean, how did that even strike you? It seems like a trial and error; I am just curios.Secondly, I have$ y = u + \frac{u}{(u^2-1)/2} + \frac{1}{(u^2-1)/2}$
$y = u + \frac{2(u +1)}{(u^2 - 1)} $

 

[Ackbach]

Hi Olok,:)

You might want to read this thread:

http://mathhelpboards.com/challenge-questions-puzzles-28/minimize-trigonometric-expression-4330.html

That's the thread I was seeking! I didn't find it, though.

 

[anemone]

Hello @anemone,

I looked at the thread and I looked a page 2, where you posted on February 2014, for the x = a - 135My question for you,

How in the world did you figure out you should let $x = 1 - 135$

I mean, how did that even strike you? It seems like a trial and error; I am just curios.

First of all, the method that you are talking about, it wasn't my solution, but I'll hazard a guess, the solver who made the substitution of $x=a-135^{\circ}$ knew it must be the correct substitution by experience! That said, practice for more math problems, chances are you will also find yourself capable to come out with a good guess.:)

Secondly, I have$ y = u + \frac{u}{(u^2-1)/2} + \frac{1}{(u^2-1)/2}$
$y = u + \frac{2(u +1)}{(u^2 - 1)} $

Good! Now, if I tell you the denominator of the term $\frac{2(u +1)}{(u^2 - 1)}$ is the difference of square, which can then be factored as $(u+1)(u-1)$, what would be your next step?

That's the thread I was seeking! I didn't find it, though.

I've to dig that out by googling it!

 

[Amad27]

First of all, the method that you are talking about, it wasn't my solution, but I'll hazard a guess, the solver who made the substitution of $x=a-135^{\circ}$ knew it must be the correct substitution by experience! That said, practice for more math problems, chances are you will also find yourself capable to come out with a good guess.:)
Good! Now, if I tell you the denominator of the term $\frac{2(u +1)}{(u^2 - 1)}$ is the difference of square, which can then be factored as $(u+1)(u-1)$, what would be your next step?
I've to dig that out by googling it!

Thank you @anemone, that is very kind of you to help moi =) I appreciate it!

We then get,

$y = u + \frac{2}{u-1}$

Since $u = sin(x) + cos(x)$ it is a composite function now.

I actually just looked back at the thread you referred me to, to see @Ackbach's solution.

It seems quite weird at this point,

How is the solution (on the page) treating $u$ as an independent variable?
It is a composite function though?

 

[MarkFL]

You still have the same function, it is just written in terms of a different variable.

Can you determine, using the second form of $u$ I gave, what your boundaries are?

 

[mathbalarka]

If $f \circ g$ is some smooth composite function formed by composition of smooth functions $f$ and $g$, minimas must occur at the points $x_0$ satisfying $t_0 = g(x_0)$ where $t_0$ is the minima of $f$, right? (convince yourself why) That's the rule we are using here.

In the case of $\displaystyle f(u) = u + \frac{2}{u-1}$, the minimas occur at $f'(u) = 0$, i.e., $u_0 = 1 \pm \sqrt{2}$. Can you determine the solutions (if any) of $f(x) = u_0$ for each of these $u_0$s? Use MarkFL's elegant form for $u(x) = \sin(x) + \cos(x)$ above.

 

[anemone]

Thank you @anemone, that is very kind of you to help moi =) I appreciate it!

We then get,

$y = u + \frac{2}{u-1}$

Hmm...there are two ways to proceed at this point:

1. We can stick to work with the variable of $u$, and solve for the minimum of $y$, or

2. We can back substitute $u = sin(x) + cos(x)$ into $y = u + \frac{2}{u-1}$ and solve for the minimum of $y$.

Can you proceed?

 

[Amad27]

Hmm...there are two ways to proceed at this point:

1. We can stick to work with the variable of $u$, and solve for the minimum of $y$, or

2. We can back substitute $u = sin(x) + cos(x)$ into $y = u + \frac{2}{u-1}$ and solve for the minimum of $y$.

Can you proceed?

Hello,

I would like to proceed with the first, but I want to see how it works first. Using some of the evidence from mathbalarka's post.

He know if we have

$H(x) = f(g(x))$ we get
$H'(x) = f'(g(x))g'(x)$

Comparing it to this, we had $u(x) = sin(x) + cos(x)$

$H(x) = f(u(x))$ where $H(x) = y$ We will get

$H'(x) = f'(u(x))(u(x))$

So my question remains, how is it possible to treat $u = sin(x) + cos(x)$ as a single variable?

From composite functions suppose we had

$f(x) = sin(x)$
$g(x) = cos(x)$
$f(g(x)) = sin(cos(x))$

The domain of $f(g(x))$ is the range of $g(x) = cos(x)$

range of $g(x) = cos(x)$ is [-1, 1]

So the domain of $f(g(x)) \implies [-1, 1]$

So in the same way, can we treat $g$ as an independent variable?

$f(g) = sin(g)$

I think it is better to work with an easier example of composition first and then leap towards difficulties.

 

[anemone]

Hello,

I would like to proceed with the first, but I want to see how it works first. Using some of the evidence from mathbalarka's post.

I'll let mathbalarka to handle your questions..:D

So my question remains, how is it possible to treat $u = sin(x) + cos(x)$ as a single variable?

We can, if we linear combined these two terms:

Are you familiar with the following:

$\sin x+\cos x=\sqrt{2}\sin(x+45^{\circ})$

That is called linear combination of sine and cosine function. You can read for more about it from this link: http://pages.pacificcoast.net/~cazelais/252/lc-trig.pdf

Back to the first method that I have mentioned, if we want to find the minimum of $y$ from the function $y=u+\dfrac{2}{u-1}$, we can easily opt for the calculus method to continue working, that is to find the first derivative of $y$ w.r.t. $u$, and then set it equals zero, and find for its corresponding $u$ value and one still need to check if that $u$ will lead to a maximum or minimum though...

Can you tell me what is your next step regarding to this method that I've outlined for you?

 

[mathbalarka]

He know if we have

$H(x) = f(g(x))$ we get
$H'(x) = f'(g(x))g'(x)$

Exactly. If $f$ and $g$ are smooth enough, then the extremums of $H(x)$ occur at $H'(x) = f'(g(x)) g'(x) = 0$, i.e., either at solutions of $f'(g(x)) = 0$ or of $g'(x) = 0$. For the former case, $g(x)$ appears as a point of extremum of $f$, that is, we are treating $g$ as an independent variable.

Comparing it to this, we had $u(x) = sin(x) + cos(x)$

$H(x) = f(u(x))$ where $H(x) = y$ We will get

$H'(x) = f'(u(x))(u(x))$

You meant $H'(x) = f'(u(x)) \bbox{u'(x)}$. Treating $u$ as an independent variable, solve $f'(t) = 0$ first for whatever $f$ you have in mind, and then solve $u(x) = t$ for $x$. In this process, you'll get $f'(u(x)) = f'(t) = 0$, hence $H'(x) = f'(u(x)) u'(x) = 0 \cdot u'(x) = 0$ and thus retrieve an extremum of $H'(x)$.

From composite functions suppose we had

$f(x) = sin(x)$
$g(x) = cos(x)$
$f(g(x)) = sin(cos(x))$

The domain of $f(g(x))$ is the range of $g(x) = cos(x)$

range of $g(x) = cos(x)$ is [-1, 1]

So the domain of $f(g(x)) \implies [-1, 1]$

So in the same way, can we treat $g$ as an independent variable?

Yes, you can. $(\sin(\cos(x)))' = -\cos(\cos(x)) \cdot \sin(x)$. Assume that the extremums occur at $\cos(\cos(x)) = 0$, solutions of which occur at $\cos(x) = n\pi/2$ for odd integers $n$. But then $n\pi/2 \notin [-1, 1]$ for odd integer $n$, so there are no solutions. Thus, the possible extremums can occur only at the solutions of $\sin(x) = 0$, i.e, $x = n\pi$ for integers $n$.

 

[Amad27]

Exactly. If $f$ and $g$ are smooth enough, then the extremums of $H(x)$ occur at $H'(x) = f'(g(x)) g'(x) = 0$, i.e., either at solutions of $f'(g(x)) = 0$ or of $g'(x) = 0$. For the former case, $g(x)$ appears as a point of extremum of $f$, that is, we are treating $g$ as an independent variable.
You meant $H'(x) = f'(u(x)) \bbox{u'(x)}$. Treating $u$ as an independent variable, solve $f'(t) = 0$ first for whatever $f$ you have in mind, and then solve $u(x) = t$ for $x$. In this process, you'll get $f'(u(x)) = f'(t) = 0$, hence $H'(x) = f'(u(x)) u'(x) = 0 \cdot u'(x) = 0$ and thus retrieve an extremum of $H'(x)$.
Yes, you can. $(\sin(\cos(x)))' = -\cos(\cos(x)) \cdot \sin(x)$. Assume that the extremums occur at $\cos(\cos(x)) = 0$, solutions of which occur at $\cos(x) = n\pi/2$ for odd integers $n$. But then $n\pi/2 \notin [-1, 1]$ for odd integer $n$, so there are no solutions. Thus, the possible extremums can occur only at the solutions of $\sin(x) = 0$, i.e, $x = n\pi$ for integers $n$.

That is very interesting, I graphed and checked that both equations do indeed have the zero at the exact same $x$. Interesting...

But the slope is never the same at any point.

Which property/rule is this of functions and composite functions?

 

[mathbalarka]

Which property/rule is this of functions and composite functions?

Not sure what you mean. It's a natural consequence of the chain rule, no?

 

[Amad27]

Not sure what you mean. It's a natural consequence of the chain rule, no?

Out of curiosity, let me ask you something,

if f(x) = sin(x) and g(x) =cos(x) then

$f(g(x)) = sin(cos(x))$

What is the domain of $f(g(x))$ ?

The answer depends on whether you choose $g$ as an independent or whether you are talking about all real, $x$.

 

[mathbalarka]

Domain of a function depends on the function. Domain of $f \circ g$ is all of $\Bbb R$ in this case but I don't understand what you're calling "domain of $f \circ g$ when $g$ is an independent variable". Do you mean : domain of $f \circ g$ when $g$ is just the identity function, or do you mean : domain of $f \circ g$ restricted to the domain of $g$?

It is essential to understand the concept of domain and range before diving in pretty advanced calculus. I'd think it'd be good for you to look at some of the introductory chapters of a calculus textbook describing these matters. :)

 

[Amad27]

Domain of a function depends on the function. Domain of $f \circ g$ is all of $\Bbb R$ in this case but I don't understand what you're calling "domain of $f \circ g$ when $g$ is an independent variable". Do you mean : domain of $f \circ g$ when $g$ is just the identity function, or do you mean : domain of $f \circ g$ restricted to the domain of $g$?

It is essential to understand the concept of domain and range before diving in pretty advanced calculus. I'd think it'd be good for you to look at some of the introductory chapters of a calculus textbook describing these matters. :)

Hello,

Here is my point/question, if you have

f(g(x))

Then the domain of $f$ is $g(x)$

Then you have a function of independent variable g

f(g) = whatever,

f'(g) set to zero will give a point where $g$ is the minimum.

Suppose f'(g) = 0 at g = 4

f(4) = $m$ the minimum value (absolute)
Then how is

the minimum of $f(x) = m$ ?

Is it a theorem of the corresponding $x$ for the correspoding $u$ and $f'(u), f'(x), g'(x), g(x)$ etc...?

 

[mathbalarka]

Hello,

Here is my point/question, if you have

f(g(x))

Then the domain of $f$ is $g(x)$

That doesn't make sense. A domain of a function is a subset of $\Bbb C$, not a function!

f(g) = whatever,

f'(g) set to zero will give a point where $g$ is the minimum.

Incorrect. $\left (f(g(x)) \right)' = f'(g(x)) g'(x)$ so $f'(g)$ set to zero will give you a point $y_0 = g(x_0)$ where $f$ has an extremum, not $g$.

Suppose f'(g) = 0 at g = 4

f(4) = $m$ the minimum value (absolute)
Then how is

the minimum of $f(x) = m$ ?

Are you asking for a proof of Fermat's theorem? First, it is not necessary that $x = 4$ will be a minimum. It'll just be an extremum.

Claim : $x = x_0$ is an extremum of $f(x)$ if and only if $f'(x_0) = 0$.

Consider the graph of $f(x)$. At $x = x_0$ there is a smooth extremum of the function. Draw a bunch of tangents near $x = x_0$, and note the change of slope that occurs at $y_0 = f(x_1)$ for $x_0 > x_1$ and $y_1 = f(x_2)$ for $x_0 < x_2$. Precisely, the angle changes from acute to obtuse, thus changing $\tan(\theta)$ from positive to negative. Hence the slope at $x = x_0$ must be zero, thus $0 = \tan(\theta) = \left [ \frac{df}{dx} \right ]_{x = x_0}$.

That's the "only if" part however. Can you prove the "if" part by yourself? Try it! :D

 

[Amad27]

That doesn't make sense. A domain of a function is a subset of $\Bbb C$, not a function!
Incorrect. $\left (f(g(x)) \right)' = f'(g(x)) g'(x)$ so $f'(g)$ set to zero will give you a point $y_0 = g(x_0)$ where $f$ has an extremum, not $g$.
Are you asking for a proof of Fermat's theorem? First, it is not necessary that $x = 4$ will be a minimum. It'll just be an extremum.

Claim : $x = x_0$ is an extremum of $f(x)$ if and only if $f'(x_0) = 0$.

Consider the graph of $f(x)$. At $x = x_0$ there is a smooth extremum of the function. Draw a bunch of tangents near $x = x_0$, and note the change of slope that occurs at $y_0 = f(x_1)$ for $x_0 > x_1$ and $y_1 = f(x_2)$ for $x_0 < x_2$. Precisely, the angle changes from acute to obtuse, thus changing $\tan(\theta)$ from positive to negative. Hence the slope at $x = x_0$ must be zero, thus $0 = \tan(\theta) = \left [ \frac{df}{dx} \right ]_{x = x_0}$.

That's the "only if" part however. Can you prove the "if" part by yourself? Try it! :D

Thanks, but I wasn't asking for a proof of Fermat's theorem (nor was that my question). My question is, (bring basic trig again),

$f(x) = cos(x)$
$g(x) = sin(x)$

$f(g(x)) = cos(sin(x))$

What is the domain of $f(g(x))$ now?

You can still put in ALL read $x$ values,

$cos(sin(x))$ is defined for all real $x$.

Now we have,

$H(x) = cos(sin(x))$ let u = sin(x)

$H(x) = cos(u)$ [still all real $x$]

$H(u) = cos(u)$ [only defined for $ -1 <= u <= 1$]

You must look at what you are finding it (with respect to).

Anyway,

New; suppose,

$f(x) = sin(x) + cos(x)$

Let $u = sin(x) + cos(x)$ to simplify things.

$f(x) = u$
$f(u) = u$

This severely changes. From a trig curve to just a line. Anyhow,

$f'(u) = 1$

Set $f'(u) = 0$, which is impossible for any $u$.

But the function $f(x) = sin(x) + cos(x)$ does have a min/max!
My point now finally is that, this method of letting the inner composite be the independent variable doesn't work (most of the time), and one must be careful with it.

Thanks

 

[mathbalarka]

My question is, (bring basic trig again),

$f(x) = cos(x)$
$g(x) = sin(x)$

$f(g(x)) = cos(sin(x))$

What is the domain of $f(g(x))$ now?

It's $\Bbb R$ : $\sin$ is defined everywhere on $\Bbb R$ and the range $\sin(\Bbb R)$ is $[1, -1]$, and $\cos$ is defined everywhere in this interval.

You can still put in ALL read $x$ values,

$cos(sin(x))$ is defined for all real $x$.

Now we have,

$H(x) = cos(sin(x))$ let u = sin(x)

$H(x) = cos(u)$ [still all real $x$]

$H(u) = cos(u)$ [only defined for $ -1 <= u <= 1$]

You have committed several mistakes in this part. 1) $H(x) = \cos(\sin(x))$, after the substitution $u = \sin(x)$, becomes $H(\arcsin(u)) = \cos(u)$, NOT $H(u) = \cos(u)$. 2) $\cos(u)$ is merely restricted to the domain of $\sin$. It's defined everywhere on $\Bbb R$.

New; suppose,

$f(x) = sin(x) + cos(x)$

Let $u = sin(x) + cos(x)$ to simplify things.

$f(x) = u$
$f(u) = u$

This severely changes. From a trig curve to just a line.Anyhow,

$f'(u) = 1$

Set $f'(u) = 0$, which is impossible for any $u$.

Again, you have made many mistakes. $f(x) = \sin(x) + \cos(x)$, after the substitution $u = \sin(x) + \cos(x)$, becomes $f(t(u)) = u$ where $t(u) = \arcsin\left(u \pm \sqrt{1-u^2}\right)$, NOT $f(u) = u$. In general, if $f(g(x)) = g(x)$ then the sub $u = g(x)$ makes $f(u) = u$. In this case, however, $f(x) = f(x)$, so the sub $u = f(x)$ will make your function $f(f^{-1}(u)) = u$ where $f^{-1}$ is the composition inverse of $f$. I'd recommend you to get used to some of these transformations and substitutions first before looking at the big picture.

My point now finally is that, this method of letting the inner composite be the independent variable doesn't work (most of the time), and one must be careful with it.

No, this method ALWAYS works, as I have proved above by application of the chain rule. It's just that you're not applying it correctly :)

 

[Amad27]

It's $\Bbb R$ : $\sin$ is defined everywhere on $\Bbb R$ and the range $\sin(\Bbb R)$ is $[1, -1]$, and $\cos$ is defined everywhere in this interval.
You have committed several mistakes in this part. 1) $H(x) = \cos(\sin(x))$, after the substitution $u = \sin(x)$, becomes $H(\arcsin(u)) = \cos(u)$, NOT $H(u) = \cos(u)$. 2) $\cos(u)$ is merely restricted to the domain of $\sin$. It's defined everywhere on $\Bbb R$.
Again, you have made many mistakes. $f(x) = \sin(x) + \cos(x)$, after the substitution $u = \sin(x) + \cos(x)$, becomes $f(t(u)) = u$ where $t(u) = \arcsin\left(u \pm \sqrt{1-u^2}\right)$, NOT $f(u) = u$. In general, if $f(g(x)) = g(x)$ then the sub $u = g(x)$ makes $f(u) = u$. In this case, however, $f(x) = f(x)$, so the sub $u = f(x)$ will make your function $f(f^{-1}(u)) = u$ where $f^{-1}$ is the composition inverse of $f$. I'd recommend you to get used to some of these transformations and substitutions first before looking at the big picture.
No, this method ALWAYS works, as I have proved above by application of the chain rule. It's just that you're not applying it correctly :)

Any links on the web where I can read about this?Now to the point, you stated,

You can't simply put in u = g(x)

So how can you convert $f(x) = u + \frac{2}{u-1}$ to
$f(u) = u + \frac{2}{u-1}$

As you said,

$f(x) \ne f(u)$ and you must find $x$ in terms of $u$.

Then how is it possible (for our example of the original problem) to convert

$f(x)$ into $f(u)$ without some transformations?

Thanks

 

[mathbalarka]

Now to the point, you stated,

You can't simply put in u = g(x)

Where? I never said that.

 

[Amad27]

Where? I never said that.

"Again, you have made many mistakes. f(x)=sin(x)+cos(x), after the substitution u=sin(x)+cos(x), becomes f(t(u))=u where t(u)=arcsin(u±1−u2‾‾‾‾‾‾√), NOT f(u)=u"

You can't just substitute in u = g(x) = sin(x) and write $u$ as the independent variable (as you suggest). So, in theory,$f(x) = u + \frac{2}{u-1}$

We use $u = sin(x) + cos(x)$, thus,

$u = \sqrt(2)\cdot sin(x + \frac{\pi}{4})$
$\frac{u}{\sqrt(2)} = sin(x + \frac{\pi}{4}$
$x + \frac{\pi}{4} = \arcsin\left({\frac{u}{\sqrt(2)}}\right)$
$x = \arcsin\left({\frac{u}{\sqrt(2)}}\right) - \frac{\pi}{4}$

So, $f(u) \ne u + \frac{2}{u-1}$

 

[mathbalarka]

You can't just substitute in u = g(x) = sin(x) and write u as the independent variable (as you suggest)

Why not? $f(\arcsin(u \pm \sqrt{1 - u^2})) = u$ works fine, as I have suggested.

EDIT : Oh, I missed the end of your post :

$f(x) = u + \frac{2}{u-1}$

We use $u = sin(x) + cos(x)$, thus,

$u = \sqrt(2)\cdot sin(x + \frac{\pi}{4})$
$\frac{u}{\sqrt(2)} = sin(x + \frac{\pi}{4}$
$x + \frac{\pi}{4} = \arcsin\left({\frac{u}{\sqrt(2)}}\right)$
$x = \arcsin\left({\frac{u}{\sqrt(2)}}\right) - \frac{\pi}{4}$

So, $f(u) \ne u + \frac{2}{u-1}$

You are correct, $f(u) \neq u + 2/(u-1)$.

 

[Amad27]

Why not? $f(\arcsin(u \pm \sqrt{1 - u^2})) = u$ works fine, as I have suggested.

EDIT : Oh, I missed the end of your post :
You are correct, $f(u) \neq u + 2/(u-1)$.

So, then how does this work? If you know

$f(u) \ne u + 2/(u-1)$

Then how is the solution reasoning

$f(u) = u + 2/(u-1)$
$f'(u) = 1 -2/(u-1)^2$? ? ? ?

Thanks

 

[mathbalarka]

So, then how does this work? If you know

$f(u) \ne u + 2/(u-1)$

Then how is the solution reasoning

$f(u) = u + 2/(u-1)$
$f'(u) = 1 -2/(u-1)^2$? ? ? ?

Thanks

Right, so this is the center of your confusions. Recall our function

$$f(x) = \sin(x) + \cos(x) + \frac1{\sin(x)} + \frac1{\cos(x)} + \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} = u(x) + \frac{2}{u(x) - 1}$$

Where $u(x) = \sin(x) + \cos(x)$. Note that $u$ ($=u(x)$) is still not considered as an independent variable : it's a function of $x$. Consider a separate function $g(t) = t + 2/(t-1)$. We see from above that $f(x) = g(u(x))$. We are considering $g(z)$ instead of $f(z)$ and differentiating $g(u)$ instead of $f(u)$. Does that make sense? e.g., $\displaystyle g'(u) = 1 - \frac2{(u-1)^2}$, not what you wrote in the post above ($f'(u) = 1 - 2/(u-1)^2$, i.e.)

 

[Amad27]

Right, so this is the center of your confusions. Recall our function

$$f(x) = \sin(x) + \cos(x) + \frac1{\sin(x)} + \frac1{\cos(x)} + \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} = u(x) + \frac{2}{u(x) - 1}$$

Where $u(x) = \sin(x) + \cos(x)$. Note that $u$ ($=u(x)$) is still not considered as an independent variable : it's a function of $x$. Consider a separate function $g(t) = t + 2/(t-1)$. We see from above that $f(x) = g(u(x))$. We are considering $g(z)$ instead of $f(z)$ and differentiating $g(u)$ instead of $f(u)$. Does that make sense? e.g., $\displaystyle g'(u) = 1 - \frac2{(u-1)^2}$, not what you wrote in the post above ($f'(u) = 1 - 2/(u-1)^2$, i.e.)

Hello @mathbalarka,

I am reading what you have written very carefully.

Oh okay, so you have a different function, $g$ instead of just $f$.

SO $g$ is a function of $x$.

One last confusion, then I'll be done (hopefully).

You still have

$g(u(x)) = f(x)$

$d/du g(u) = g'(u)$
$d/du g(u) \ne f'(x)$
Because $f'(x) = d/dx f(x)$

Wait let me get a fact straight.

Is g(u(x)) a function of $u$ or a function of $x$??

Thanks Balarka