If I want to solve $$\frac{d}{d\frac{1}{x}} x$$ is my approach correct? As $$\begin{align*} \frac{d}{d\frac{1}{x}}x&=\\ \text{with }\frac{1}{x}&=y\\ \frac{d}{dy}\frac{1}{y}&=-\frac{1}{y^2}\\ &=-\frac{1}{\left(\frac{1}{x}\right)^2}\\ &=-x^2 \end{align*}$$ Is this approach correct, or did I miss something?
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1Your solution is accurate. – Galc127 Feb 08 '16 at 10:04
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http://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio – Alex Feb 08 '16 at 10:23
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$\frac{\mathrm{d} x}{\mathrm{d} \frac{1}{x}}=\left (\frac{\mathrm{d} \frac{1}{x}}{\mathrm{d} x} \right )^{-1}=\left ( \frac{-1}{x^{2}} \right )^{-1}=-x^{2}$
John11
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$x=(\frac{1}{x})^{-1}$. Set $y=\frac{1}{x}$ and use ordinary Differentiation rules. Then express $y$ in Terms of $x$.
kryomaxim
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