Determine dy/dx of the following and simplify if possible

[DevonZA]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

cos(x-y)=ysinx

My attempt:
-sin(x-y(x'-y')=y'cosx.x'

Yeah I'm stuck.. I know it is differentiation of implicit functions and I need to make y' the subject of the formula.

 

[ShayanJ]

Do you know that $x'=\frac{dx}{dx}=1$?

 

[DevonZA]

Do you know that $x'=\frac{dx}{dx}=1$?

I do now

 

[ShayanJ]

Actually you've done something wrong. You should have $(y'-1)\sin(x-y)=y'\sin x+y \cos x \Rightarrow [\sin(x-y)-\sin x]y'=y\cos x +\sin(x-y)$.
Can you continue?

 

[DevonZA]

y'= $\frac{ycosx+sin(x-y)}{sin(x-y)-sinx}$

y'=$\frac{ycosx}{-sinx}$

Therefore $\frac{dy}{dx}$ = $\frac{ycosx}{-sinx}$

 

[ShayanJ]

What happened to $\sin(x-y)$? You can't do that!

 

[DevonZA]

What happened to $\sin(x-y)$? You can't do that!

Um it got canceled because it was the numerator and denominator of the fraction? My bad

 

[ShayanJ]

Um it got canceled because it was the numerator and denominator of the fraction? My bad

If it was something like $\frac{f(x)g(x)}{f(x)h(x)}$, then you could cancel f(x) and have $\frac{g(x)}{h(x)}$. But here you have $\frac{f(x)+g(x)}{f(x)+h(x)}$. you can't simplify further.

 

[DevonZA]

$\frac{dy}{dx}$ = $\frac{ycosx+sin(x-y)}{sin(x-y)-sinx}$

Is the final answer then?

 

[ShayanJ]

Yes

 

[DevonZA]

Thank you very much

 

[DevonZA]

Final answer attached. Thanks to all who helped.