I have this implicit function $$y=f(x) \iff \sin(x+y)=k \sin(x), \quad$$ where $k>1$ is a constant.

I would like to know how a small variation in $x$ propagates on $y$.

I think I need to do an implicit differentiation but then it is not so clear to me how to solve the problem.

So the derivative of the LHS is

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

and the derivative of the RHS is

$$\frac{\mathrm{d}}{\mathrm{d}x}(k\sin(x)) = k\cos(x)$$

And solving for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$ gives


and now how can I continue?

Thank you.

  • 1
    $\begingroup$ How about just rewriting this as $y=\sin^{-1}(k\sin(x)) - x$ and differentiating the right side? $\endgroup$ – Thomas Andrews Jul 6 '11 at 15:55
  • $\begingroup$ @Thomas Andrews I do not know how to explain it but I am not so comfortable with the inverse trigonometric functions... :-) $\endgroup$ – Alessandro Jacopson Jul 6 '11 at 16:05

It depends on the result you would like to obtain. First of all, $$ dy = \left(\frac{k\cos{x}}{\cos(x+y)}-1\right)dx $$ so the propagation of $dx$ on $dy$ depends on $x$ as well as on $y(x)$. You can make better if recall that $\cos^2(x+y) + k^2\sin^2x = 1$ which can help you to rewrite the denominator and obtain the dependence only on $x$.

  • $\begingroup$ +1, thank you, it was the dependency on y(x) that puzzles me a bit. $\endgroup$ – Alessandro Jacopson Jul 6 '11 at 9:31
  • $\begingroup$ @uvts_cvs: note that in general this trick will not work and you will have $dy = f(x,y)dx$. What puzzled you? $\endgroup$ – Ilya Jul 6 '11 at 9:41
  • $\begingroup$ I was puzzled because I was not expecting the full dependency on $x$ and $y(x)$ but you are obviously right. $\endgroup$ – Alessandro Jacopson Jul 6 '11 at 9:47
  • $\begingroup$ Let's say I am interested at $x=0$ where we have $y=0$. How can I plug in these numbers into the $\frac{dy}{dx}$? I get $\frac{dy}{dx}=k-1$ and now how can I interpret this result? A small (how much small?) change in $x$ will be amplified by $k-1$? $\endgroup$ – Alessandro Jacopson Jul 6 '11 at 16:10
  • $\begingroup$ I have an advise for you to open $\sin(x+y)$, denote $z = \sin y$ and find $z(x)$ if you like to deal with explicit dependencies. Note that for $x=0$ you have $y = \pi k$. $\endgroup$ – Ilya Jul 6 '11 at 16:25

Rewrite your equation as:


Then notice that:

$$\cos(x+y) = \pm\sqrt{1-\sin^2(x+y)} = \pm \sqrt{1-k^2\sin^2(x)}$$


$$\frac{\mathrm{d}y}{\mathrm{d}x}=\pm \frac{k\cos(x)}{\sqrt{1-k^2\sin^2(x)}}-1$$

Where the sign is determined by the sign of $\cos(x+y)$.


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