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The main question being asked: is it possible an $f(x)$ which satisfies the below equation where $u=u(x)$;if so, how?

$$\boxed{ \frac{df(u^2)}{d(u^2)} = \left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2} \quad ,u = u(x)\qquad(*)}$$ or similarly after using the transformation in the Remarks section: $$\boxed{\frac{df(u^2)}{du} = 2u\left[\left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2}\right] \quad ,u = u(x) \qquad (**)}$$


Part 1:

This stems from the initial question assigned to me which was:

Find an $f(x)$ which satisfies: $$ f'(\sin^2x) = \cos^2x + \tan^2x, \quad 0<x<1$$

(I have left this part in 'prime' notation because this is how the original question was presented.)

The proper solution involves the following method:

$$ f'(\sin^2x) = (1-\sin^2x) + \frac{\sin^2 x}{1-\sin^2 x}$$

Let $u = \sin^2x$, then:

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

$$ f(u) = \int\,\left[(1 - u) + \frac{u}{1-u} \right] du$$

$$ f(u) = \int\,\left[(1 - u) + \frac{u-1+1}{1-u} \right] du$$

$$ = \int\,\left[-u + \frac{1}{1-u} \right] du$$

$$ \therefore f(x) = -\frac{1}{2}u^2-\ln|1-u|=-\frac{1}{2}x^2-\ln|1-x|$$


Part 2:

My question is if this can be solved with the following alternate method:

$$ \frac{d(f(\sin^2x))}{d(\sin^2x)} = \cos^2x + \tan^2x, \quad 0<x<1$$

then,

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

Let $u = \sin x$, $du = \cos x\,dx$ then:

$$ \frac{d(f(u^2))}{d(u^2)} = \left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2} \qquad$$

Can this equation be solved to find an $f(x)$ which satisfies this relation regardless of what $u(x)$ actually is?

See Main Edit 1&2 for clarification on why it is in this form

The following steps are INCORRECT, as this is not how the chain rule works. But there is something in my head grinding such that I feel that it can be solved in a way invoking the chain rule but I can't get my thoughts wrapped around it properly.

(I have left this part in 'prime' notation because this displays my erroneous thought process of misunderstanding differentials which probably led me to the wrong answer.)

$$ 2u\, f'(u) = \left(u'\right)^2 +\frac{u^2}{\left(u'\right)^2}$$

Clearly $ f'(u^2) \ne 2u\, f'(u)$, however continuing with this incorrect thought..

$$ f'(u) = \frac{1}{2} \left[\frac{\left(u'\right)^2}{u}+\frac{u}{u'}\right]$$

$$ f(u) = \int\frac{1}{2} \left[\frac{\left(u'\right)^2}{u}+\frac{u}{u'}\right] $$

Is there anyway I can actually correctly use the chain rule starting from $(*)$ to reach this point and then integrate, or solve it through some method for differential equations? I've never experienced a question with the square of a derivative so I was wondering if someone could give me insight on this.


Main Edit 1: As @JJacquelin noted, the notation with 'prime' causes quite problem with what the question being asked is:

Using the fact that $u = \sin x\,\, \text{is a function of}\, x$

The line:

$$ f'(u^2) = \left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2} \qquad$$

can be written as such:

$$ \frac{df(u^2)}{d(u^2)} = \left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2} \quad ,u = u(x) \qquad(*)$$

And as so, I have replaced all appropriate parts of this post with Leibniz notation to suggest clarity.

Now can I find an $f(x)$ which satisfies this equation?

I have left my previous work as is, so that I may reference it one day in case I come across a problem of pinpointing what my prime notation means.

Please do let me know if there are any further clarifications that need to be made before this can be solved properly.


Edit 2: Proof of why [incorrectly] believed it is $\frac{d}{dx}$:

Thank you to @JJacquelin once again for helping me clarify the differentials in the question. I have edited the general question with reflection to their analysis, and hope to recieve the answer I was looking for now that the notion of differentials is cleared up. I was incorrect in my reasoning before and it is actually $\frac{df(X)}{dX}$ for any dummy variable $X$, contrary to what I had thought as you can see below.:

If we start from the solution of Part 1:

$$f(x) = -\frac{1}{2}x^2-\ln|1-x|$$

$$\frac{df(x)}{dx} = -x+\frac{1}{1-x}$$ $$\frac{df(x)}{dx} = -x+\frac{1-x+x}{1-x}$$ $$\frac{df(x)}{dx} = -x+\frac{1-x}{1-x}+\frac{x}{1-x}$$ $$\frac{df(x)}{dx} = -x + 1 + \frac{x}{1-x}$$ $$\frac{df(x)}{dx} = 1 - x +\frac{x}{1-x}$$

Now $x \mapsto \sin^2x$:

$$\frac{df(\sin^2x)}{dx} = 1 -\sin^2x+\frac{\sin^2x}{1-\sin^2x}$$ $$\frac{df(\sin^2x)}{dx} = \cos^2x + \tan^2x$$ But this turns out to be wrong, when $x\mapsto \sin^2 x$ it becomes: $$\frac{df(\sin^2x)}{d(\sin^2x)} = \cos^2x + \tan^2x$$ $$f'(\sin^2x)=\cos^2x + \tan^2x$$ Which is what we started with. Now I am beginning to doubt my self, because I am not sure if the $dx$ changes to $d(\sin^2x)$ when $x\mapsto\sin^2x$. I would appreciate it if someone could help me realize which one it is, because then I can't begin to even grasp the main question without formulating the correct statement.

Thank you for @JJacquelin clearing this up.


Remarks: On another note, I looked back to some of my differential work from university and came across a homework problem we proved which shows that:

$$\frac{d^2x}{dy^2}= -\frac{\frac{d^2y}{dx^2}}{\left(\frac{dy}{dx}\right)^3}$$

Which leads me to wonder, can the terms $\left(\frac{du}{dx}\right)^2$ be transformed into some $n$-th derivative? More generally, are there possible transformations to convert powers of derivatives to $n$-th derivatives in order to simplify this problem into a differential equation which can be solved using usual methods?

Also something @JJacquelin helped me realize about use of the chain rule, which may be of some help to progress:

$$\frac{df(u^2)}{d(u^2)}= \frac{df(u^2)}{2udu}$$

which may simplify the original question to:

$$\frac{df(u^2)}{du} = 2u\left[\left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2}\right] \quad ,u = u(x) \qquad (**)$$

These two identities might be of some help...possibly.

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  • $\begingroup$ Sorry to be clumsy, but it appears still ambiguous for me. For example does $f'(\sin^2(x)$ means $\frac{d}{d(\sin^2(x))}f(\sin^2(x))$ or $\frac{d}{dx}f(\sin^2(x))$ ? In the wording, they are still a lot of primes. Excuse me to repeat, in this kind of problem, I think that the symbol prime should be suppressed everywhere. $\endgroup$ – JJacquelin Feb 16 '17 at 22:50
  • $\begingroup$ @JJacquelin I have changed the primes into Leibniz notation in every part of the question except for Part 1, because that was how the original question was presented to me on a work sheet, and in the second half of part 2 because I want to make clear my erroneous work that might have stemmed from using prime notation. Everywhere else however, the prime notation has been removed and replaced with d/dx, not d/d(sin^2x) because if you take the solution $f(x)$ from part 1 and then take $df(x)/dx$ of that function and then substitute $\sin^2x\, \text{for}\, x$ you will get the original question. $\endgroup$ – Hushus46 Feb 17 '17 at 14:10
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In fact, the preceding posts are more comments than answers. The aim was to clarify the problem. After discussion, this was done. So, these preliminary posts could be deleted.

In fact, the question is :

Is it possible an $f$ which satisfies the below equation where $u=u(x)$;if so, how? $$\boxed{ \frac{df(u^2)}{d(u^2)} = \left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2} \quad ,u = u(x)}$$ From what was discussed before : $$f(u^2)=\int \left( \left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2} \right)2udu$$ One can take any differentiable function $u(x)$ and put it into the above integral. Doesn't mater if there is, or not, a closed form for the integral. This proves that the function $f$ exists (fonction defined by an integral).

Let $F(x)$ a given function. With $u(x)=F(x)$ the derivative $\frac{du}{dx}=F'(x)$ is known.

In the next expression $$\left( \left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2} \right)2udu=\left( \left(F'(x)\right))^2 + \frac{ \left(F(x)\right)^2}{ \left(F'(x)\right)^2 } \right)2F(x)F'(x)dx$$ all the functions are known.

Thus the function $I(x)$ defined by the next integral is known (even if no closed form can be expressed) :

$$I(x)=\int \left( \left(F'(x)\right))^2 + \frac{ \left(F(x)\right)^2}{ \left(F'(x)\right)^2 } \right)2F(x)F'(x)dx$$ $$f(u^2)=I(x)$$ The inverse function of $u=F(x)$ is $x=F^{-1}(u)$ $$f(u^2)=I\left(F^{-1}(u)\right)$$ With $X=u^2$ $$\boxed{ f(X)=I\left(F^{-1}(X^{1/2})\right)}$$ This shows how the function $f(X)$ can be obtained.

EXAMPLES :

This was already done in case of $F(x)=\sin(x) \quad\to\quad f(X)=-\frac{1}{2}X^2-\ln|1-X|$.

We will do it with another function, for example the very simple $F(x)=x$

$F'(x)=1 \quad\to\quad I(x)=\int \left( 1 + x^2 \right)2xdx = x^2+\frac{2}{3}x^3$

$x=F^{-1}{u}=u=X^{1/2} \quad\to\quad F(X)=I\left(F^{-1}(X^{1/2})\right)=(X^{1/2})^2+\frac{2}{3}(X^{1/2})^3$

$F(X)=X+\frac{2}{3}X^{3/2}$

OTHER EXAMPLE :

More complicated, with $F(x)=\ln(x)$

$F'(x)=\frac{1}{x}$

$I(x)=\int \left( \frac{1}{x^2} + \frac{ \left(\ln(x)\right)^2}{ \frac{1}{x^2} } \right)2\ln(x)\frac{1}{x}dx $

$I(x)= x^2\ln^3(x)-\frac{3}{2}x^2\ln^2(x)+\frac{3}{2}x^2\ln(x)-\frac{\ln(x)}{x^2}-\frac{3}{4}x^2-\frac{1}{x^2}$

$x=F^{-1}\left(u\right)=e^u=e^{X^{1/2}}$ $$f(X)=e^X X^{3/2}-\frac{3}{2}e^X X+\frac{3}{2}e^X X^{1/2}-\frac{X^{1/2}}{e^X}-\frac{3}{4}e^X-\frac{1}{e^X}$$

As already pointed out, a closed form for $f(X)$ cannot be derived in all cases. This requires that closed forms be known for the integral and for the inverse function.

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  • $\begingroup$ Wow, that was a very interesting approach. This is the answer I was looking for, and it helped me understand a lot. Sorry for the confusion beforehand in clarifying the question, I appreciate all the work you put into helping me. $\endgroup$ – Hushus46 Feb 19 '17 at 13:26
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The trouble comes from the ambiguity of the symbol prime, which doesn't specifies with respect to which variable the differentiation is done.

Does $\quad f'(u)= \begin{cases} \frac{d}{du} f(u)\qquad(1)\\ \text{or} \\ \frac{d}{dx}f(u) \qquad(2) \end{cases}\quad $ ?

Does $\quad f'(\sin^2(x))= \begin{cases} \frac{d}{d(\sin^2(x))} f(\sin^2(x))\qquad(1)\\ \text{or} \\ \frac{d}{dx}f(\sin^2(x)) \qquad(2) \end{cases}\quad $ ?

It seems that, in writing $\quad f'(\sin^2x) = (1-\sin^2x) + \frac{\sin^2 x}{1-\sin^2 x}\quad$ you understand "prime" in sens of definition $(2)$ ,

while in writing $\quad f'(u) = (1 - u) + \frac{u}{1-u}\quad$ you understand "prime" in sens of definition $(1)$.

Both are different and leads to different further calculus. So, it is difficult to give a definitive answer to your question.

When two different variables, sometimes $u$ sometimes $x$, are involves in differentiations, it should be better to avoid the symbol prime. I suggest to edit your question with no symbol prime but with symbols $d$, in order to make it not ambiguous.

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  • $\begingroup$ Indeed, a valid point that I should've noticed earlier. I am currently going to bed right now, but I will edit it as soon as I am up and awake, and hope to further receive a definitive answer. Thank you for the apt response :) $\endgroup$ – Hushus46 Feb 16 '17 at 6:34
  • $\begingroup$ I have added a main edit in reference to your post, I hope this clears up the confusion with the prime notation. Thank you for letting me know about this clarity issue. $\endgroup$ – Hushus46 Feb 16 '17 at 14:33
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Since the original question was presented on the form : $$f'\left(\sin^2(x)\right)=\cos^2(x)+\tan^2(x)$$ probably the symbol prime means $$f'(X)=\frac{df(X)}{dX}\text{ , for any dummy variable }X.$$ So, with $X=\sin^2(x)\quad\to\quad f'\left(\sin^2(x)\right)=\frac{df\left(\sin^2(x)\right)}{d\left(\sin^2(x)\right)}=\cos^2(x)+\tan^2(x)$

With this interpretation, your first part is correct because, with symbol $X=u=\sin^2(x)$ the equation to be solved is $$f'(u)=\frac{df(u)}{du}=(1-u)+\frac{u}{1-u}$$ $$f(u)=\int \left((1-u)+\frac{u}{1-u}\right)du$$ leading to $$f(u)=-\frac{1}{2}u^2-\ln|1-u|+C \quad\to\quad f(x)=-\frac{1}{2}x^2-\ln|1-x|+C$$

SECOND PART, with change of variable $v=\sin(x)$

I don't use the same symbol $u$ in order to avoid confusion.

The equation to be solved is $$f'(v^2)=\frac{df(v^2)}{d(v^2)}=(1-v^2)+\frac{v^2}{1-v^2}$$ $\frac{df(v^2)}{d(v^2)}=\frac{df(v^2)}{2vdv}=(1-v^2)+\frac{v^2}{1-v^2}$

$\frac{df(v^2)}{dv}=2v\left((1-v^2)+\frac{v^2}{1-v^2}\right)$ $$df(v^2)=\left((2v-2v^3)+\frac{2v^3}{1-v^2}\right)dv$$ $$f(v^2)=\int\left(2v-2v^3+\frac{2v^3}{1-v^2}\right)dv$$ $$f(v^2)=-\frac{1}{2}v^4-\ln|1-v^2|+c$$ As expected, the result is the same as above : $\quad f(x)=-\frac{1}{2}x^2-\ln|1-x|+c$

NOTE :

It is possible to solve the original equation without any change of variable : $$f'\left(\sin^2(x)\right)=\frac{df\left(\sin^2(x)\right)}{d\left(\sin^2(x)\right)}=\cos^2(x)+\tan^2(x)$$

$$f'\left(\sin^2(x)\right)=\frac{df\left(\sin^2(x)\right)}{2\sin(x)\cos(x)dx}=\cos^2(x)+\tan^2(x)$$

$$df=2\sin(x)\cos(x)\left(\cos^2(x)+\tan^2(x)\right)dx$$

$$f=\int df=\int 2\sin(x)\cos(x)\left(\cos^2(x)+\tan^2(x) \right)dx$$ $\int 2\sin(x)\cos(x)\left(\cos^2(x)+\tan^2(x) \right)dx=-\frac{1}{2}\sin^4(x)-\ln(\cos^2(x))+c$ $$f\left(\sin^2(x)\right)=-\frac{1}{2}\sin^4(x)-\ln|\cos^2(x)|+c$$

$$f\left(x\right)=-\frac{1}{2}x^2-\ln|1-x^2|+c$$

As expected, this is the same result as above.

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  • $\begingroup$ Thank you for the lengthy and detailed answer on showing me what the prime symbol actually means in this question, and even showing me how to do it without any transformations. However this is only helps me clarify part of the big question, which is solving the differential equation without knowing that the first term on the R.H.S can be replaced with $1-v^2$ but rather $\left(\frac{dv}{dx}\right)^2$. I will edit my post to reflect your clarity, and I really appreciate your efforts in helping with this so far. $\endgroup$ – Hushus46 Feb 18 '17 at 12:35
  • $\begingroup$ So far, I cannot understand because you wrote : << This stems from the initial question assigned to me which was: Find an $f(x)$ which satisfies: $$ f'(\sin^2x) = \cos^2x + \tan^2x, \quad 0<x<1$$ >> Putting $v=\sin(x)$ into it gives $=1-v^2+\frac{v^2}{1-v^2}$ . $\endgroup$ – JJacquelin Feb 18 '17 at 13:46
  • $\begingroup$ Yes putting $v=\sin(x)$ results to that. However, when I saw the question initially I didn't think of using $\sin^2x+\cos^2x=1$ identity. I saw that the derivative of $\sin(x)$ was $\cos(x)$ and $\tan(x)$ can also be represented by $\sin(x)$ and $\cos(x)$ so I replaced everything with sin and its derivative. $$f'(\sin^2x) = \cos^2x + \tan^2x, \quad 0<x<1$$ $$f'(u^2)= \left(\frac{du}{dx}\right)^2 + \frac{u^2}{\left(\frac{du}{dx}\right)^2}$$ And that's what I'm looking to see if solvable. To find an $f(x)$ which satisfies this equation regardless of what u(x) is. $\endgroup$ – Hushus46 Feb 18 '17 at 13:49
  • $\begingroup$ I only added part 1 about the initial question to show where this all started from, but now I'm not looking to solve it when $u(x) = \sin(x)$. Part 2 is what I really am looking for, while part 1 is the answer that the professor posted online and when I first looked at that question I didn't think of using that transformation to $1-v^2$ but rather that the derivative of $\sin(x)$ is $\cos(x)$ and going on from there. I hope this makes more sense, but if you want to talk in a chat room to really understand I am okay with that as well. $\endgroup$ – Hushus46 Feb 18 '17 at 13:55
  • $\begingroup$ I added the detailed steps at the end of my previous answer. Is it OK ? In fact, I am a bit lost about what you need exactly. If you have more questions, please, edit them in the comments, not in the above main question which is now very long and too confusing for me. $\endgroup$ – JJacquelin Feb 18 '17 at 18:51

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