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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

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{k\cos(x)-\cos(x+y)}{\cos(x+y)}$

and now how can I continue?

Thank you.

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    @Thomas Andrews I do not know how to explain it but I am not so comfortable with the inverse trigonometric functions... :-)2011-07-06

2 Answers 2

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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 $x$.

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    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 $y = \pi k$.2011-07-06
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Rewrite your equation as:

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

Then notice that:

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

So:

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

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