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!

#Calculus #Trigonometry #ImplicitDifferentiation

Easy Math Editor

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Comments

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.

Siddhartha Srivastava - 6 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$