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# If tan(xy) = xy, show that y' = -y/x.

If $$\tan(xy) = xy$$, show that $$y' = -y/x$$.

These are my steps : $$xy'+y = \sec^2(xy) \times (xy'+y)$$) which implies $$\sec^2(xy)=1$$ [ $$(xy'+y)$$ cancels out on both sides.]

And then I don't know how to proceed. Help!

Note by Tanvi Bhakta
3 years, 4 months ago

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You did it correctly. You just made the mistake of cancelling out the $$xy' + y$$ term . Remember you can't cancel out terms unless you're sure they're not zero.

So you've got $$xy' + y = \sec^2(xy) \times (xy' + y)$$

$$(xy' + y)(\sec^2(xy) - 1) = 0$$

So this implies $$(\sec^2(xy) - 1) = 0$$ or $$(xy' + y) = 0$$

Case 1 - $$\sec^2(xy) = 1$$ implies $$\tan(xy)=xy = 0$$. This implies $$x = 0$$ or $$y =0$$.

If $$x = 0$$, then you can't prove it.

If $$y = 0$$, then $$y' = 0 = \dfrac{-y}{x}$$.

Case 2 -$$(xy' + y) = 0 \implies xy' = - y \implies y' = \dfrac{-y}{x}$$ which is what you wanted to prove.

- 3 years, 4 months ago