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Abhisar

  • one year ago

Can some body make me understand this? \[\frac{d T}{d (1/T)} = \frac{1}{\left(\frac{d (1/T)}{d T}\right)} \] How?

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  1. Abhisar
    • one year ago
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    @ganeshie8 @myininaya

  2. Elsa213
    • one year ago
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    @Hero

  3. myininaya
    • one year ago
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    \[\frac{dT}{d(\frac{1}{T})} \\ \text{ \let } u=\frac{1}{T} \text{ then } T=\frac{1}{u} \\ \text{ so } \frac{dT}{d(\frac{1}{T})} =\frac{d(\frac{1}{u})}{du} \cdot \frac{du}{dT}\]

  4. myininaya
    • one year ago
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    \[\frac{1}{\frac{d(\frac{1}{T})}{dT}}=\frac{dT}{d(\frac{1}{T})} \text{ since the reciprocal of } \frac{d(\frac{1}{T})}{dT} \text{ is } \frac{dT}{d(\frac{1}{T})}\]

  5. myininaya
    • one year ago
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    wait what do you want to understand about this?

  6. Abhisar
    • one year ago
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    Like how can you write \[\frac{d T}{d (1/T)}~as~\frac{1}{\left(\frac{d (1/T)}{d T}\right)} \]

  7. Abhisar
    • one year ago
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    Have you used chain rule or something?

  8. myininaya
    • one year ago
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    yes I used chain rule above

  9. Abhisar
    • one year ago
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    Umm..But I thought chain rule is used when we differentiate a function with respect to different function. Right?

  10. myininaya
    • one year ago
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    well I did introduce a new function

  11. myininaya
    • one year ago
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    would you agree that: \[\frac{a}{b}=\frac{1}{\frac{b}{a}}?\]

  12. Abhisar
    • one year ago
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    Yes.

  13. Abhisar
    • one year ago
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    Whoops!!!!!!! lol, I was thinking it to be some complex stuff.

  14. Abhisar
    • one year ago
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    Thanks a bunch c:

  15. myininaya
    • one year ago
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    np

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