Calculus Challenge

[joigus]

Differentiate:

\[ f\left(x\right)=\log_{x}\left(\sin x\right) \]

Solution:

Spoiler

\[ y=\log_{x}\left(\sin x\right) \]

\[ x^{y}=\sin x \]

\[ e^{y\ln x}=\sin x \]

\[ e^{y\ln x}\left(y'\ln x+\frac{y}{x}\right)=\cos x \]

\[ y'\ln x+\frac{y}{x}=e^{-y\ln x}\cos x \]

\[ y'\ln x=e^{-y\ln x}\cos x-\frac{y}{x} \]

\[ y'=\frac{1}{\ln x}\left(e^{-y\ln x}\cos x-\frac{y}{x}\right) \]

\[ y'=\frac{1}{\ln x}\left(e^{-\ln x\log_{x}\left(\sin x\right)}\cos x-\frac{\log_{x}\left(\sin x\right)}{x}\right) \]

Simplifications:

\[ e^{-\ln x\log_{x}\left(\sin x\right)}=\left(e^{\ln x}\right)^{-\log_{x}\left(\sin x\right)}=x^{-\log_{x}\left(\sin x\right)}=\frac{1}{x^{\log_{x}\left(\sin x\right)}}=\frac{1}{\sin x} \]

With all this,

\[ y'=\frac{1}{\ln x}\left(\cot x-\frac{\log_{x}\left(\sin x\right)}{x}\right) \]

I hope I didn't make any silly mistake. If you find any other fun way to solve it, please let me know.

[joigus]

A final embelishment:

Spoiler

Change of base for logs:

\[ \log_{x}b=\frac{\log_{a'}b}{\log_{a'}x} \]

Final (simplified) answer in terms of elementary functions:

\[ y'=\frac{1}{\ln x}\left(\cot x-\frac{\ln\left(\sin x\right)}{x\ln x}\right) \]

 

[NTuft]

change of base, 

Spoiler

 

quotient rule: 

chain rule...

appreciated your initial manipulation once I saw it.

Edited September 10, 2022 by NTuft

[NTuft]

On 8/30/2022 at 3:18 AM, joigus said:

I hope I didn't make any silly mistake.

Same goes for me here:

Spoiler

 : 

  :  

also from change of base:  so log_xsin(x)*ln(x)=ln(sinx)
f(x)=a^x=(e^lna)^x=e^xlna
f'(x)=e^xlna(lna)=a^xlna

d/dx(e^lnsin(x))=e^lnsin(x)(ln(sinx))

whereas you go towards e^yln(x)(y'lnx + y/x)
I think the y/x term then contributes to a wrong result?...

[edit] but, looking at what I've got now
e ^lnsin(x)lnsin(x)=cosx  divide both sides by lnx to recover

e^lnsin(x)y=cos(x)/ln(x)

y=cos(x)/ln(x)*e^lnsin(x) = cos(x)/ln(x)*sin(x) = cot(x)/ln(x) and I'm not sure that's going in the right direction...

Trying to solve:

Spoiler

apply quotient rule to find y'= 

 

Edited September 11, 2022 by NTuft

[NTuft]

e ^lnsin(x)lnsin(x)=cosx  divide both sides by lnx to recover y

e^lnsin(x)y=cos(x)/ln(x)

(since e^lnsin(x)=sin(x))

(sin(x)/ln(x))y=cos(x)/ln(x), times both sides by ln(x)/sin(x) and

y=cot(x) which is wrong.

back to prior try, I think chain rule gets towards it but I don't quite have it and I'll spare any spoilers.

 

[joigus]

11 minutes ago, NTuft said:

back to prior try, I think chain rule gets towards it but I don't quite have it and I'll spare any spoilers.

You're on the right track. You may be just a sign/and a power off? I don't have the time to check it now, but you got the main 2 ideas for solving it.

Yes, please don't spoil the fun in case people want to give it a try.  

It's not a standard derivation technique because the function is so weird, but I'm sure someone's solved it somewhere else.

[NTuft]

If the c.o.b. I made is valid Wolfram Alpha does not agree with answer. Please demonstrate in maths or words you are boss over Wolfram Alpha, I believe it. Still potential problem with y/x term's appearance but I do not conclude you are wrong. I want to boss Wolfram Alpha around, but I need @Sensei or grasshopper to write computer code for it.

Thanks

[joigus]

7 hours ago, NTuft said:

If the c.o.b. I made is valid Wolfram Alpha does not agree with answer. Please demonstrate in maths or words you are boss over Wolfram Alpha, I believe it. Still potential problem with y/x term's appearance but I do not conclude you are wrong. I want to boss Wolfram Alpha around, but I need @Sensei or grasshopper to write computer code for it.

Thanks

I just asked WA, and totally agrees with my calculation. I can PM you solution from "Fram" if you want.

Cheers

[NTuft]

Oi, good on you. I'll re-visit it and try to see where I was mistaken. Thanks for the challenge!

[joigus]

1 hour ago, NTuft said:

Oi, good on you. I'll re-visit it and try to see where I was mistaken. Thanks for the challenge!

You're very welcome!

[joigus]

Here's the Wolfram Alpha output that --as I told you-- agrees with my calculation. But you will find no clue as to why it's correct.

Spoiler

Cheers.