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In one example of my textbook I see that it uses $y\frac{dy}{dt} = \frac{1}{2}\frac{d(y^2)}{dt}$. But I am not quite sure where it comes from. It seems like it is related to integrating $y$ because the existence of $\frac{1}{2}$ and $y^2$ on the right hand side. Please show me how this equality is true.

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  • $\begingroup$ think of chainrule $\endgroup$
    – imranfat
    Oct 30, 2017 at 1:27
  • $\begingroup$ by chain rule... look at right hand side and apply chain rule there $\endgroup$
    – randomgirl
    Oct 30, 2017 at 1:27

1 Answer 1

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$$\frac12 \frac{\mathrm d(y^2)} {\mathrm dt} = \frac12 \frac{\mathrm d(y^2)} {\mathrm dy} \frac{\mathrm dy} {\mathrm dt} = \frac12 (2y) \frac{\mathrm dy} {\mathrm dt} = y \frac{\mathrm dy} {\mathrm dt}$$

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