$xy=1$ can be written as $y=\frac{1}{x}$. The derivative for $\frac{1}{x}$ is $-\frac{1}{x^2}$
But if I were to implicitly differentiate the original function $xy=1$, I get: $$\frac{dy}{dx} = -\frac{y}{x}$$ Where did I go wrong?
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5$\begingroup$ Subsitute $y=\frac{1}{x}$ into what you have. $\endgroup$– PepperSausageJan 2, 2017 at 16:09
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$\begingroup$ you mean in the numerator instead of y? $\endgroup$– user3507767Jan 2, 2017 at 16:12
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1$\begingroup$ $- \frac{y}{x} = - \frac{y \, x}{x^2} = - \frac{1}{x^2}$ $\endgroup$– Hirek KubicaJan 2, 2017 at 16:20
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2 Answers
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You have the right answer, just substitute your $y(x)$:
$$\frac{dy}{dx} = -\frac{y}{x}=-\frac{(\frac{1}{x})}{x}=-\frac{1}{x}\times\frac{1}{x}=-\frac{1}{x^2}= -y^2$$
answered Jan 2, 2017 at 16:13
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By the product rule we get:
$$(xy)'=1'\Rightarrow x'y+y'x=0 \Rightarrow y+y'x=0 \Rightarrow y'=\frac{-y}{x} \tag{1}$$
But $\displaystyle xy=1\Rightarrow y=\frac{1}{x}$, replacing in (1), we get: $$y'=\frac{-1}{x^2}$$
answered Jan 2, 2017 at 22:13