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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?
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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myininaya
 one year ago
Best ResponseYou've already chosen the best response.4\[\frac{dT}{d(\frac{1}{T})} \\ \text{ \let } u=\frac{1}{T} \text{ then } T=\frac{1}{u} \\ \text{ so } \frac{dT}{d(\frac{1}{T})} =\frac{d(\frac{1}{u})}{du} \cdot \frac{du}{dT}\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.4\[\frac{1}{\frac{d(\frac{1}{T})}{dT}}=\frac{dT}{d(\frac{1}{T})} \text{ since the reciprocal of } \frac{d(\frac{1}{T})}{dT} \text{ is } \frac{dT}{d(\frac{1}{T})}\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.4wait what do you want to understand about this?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Like how can you write \[\frac{d T}{d (1/T)}~as~\frac{1}{\left(\frac{d (1/T)}{d T}\right)} \]

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Have you used chain rule or something?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.4yes I used chain rule above

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Umm..But I thought chain rule is used when we differentiate a function with respect to different function. Right?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.4well I did introduce a new function

myininaya
 one year ago
Best ResponseYou've already chosen the best response.4would you agree that: \[\frac{a}{b}=\frac{1}{\frac{b}{a}}?\]

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Whoops!!!!!!! lol, I was thinking it to be some complex stuff.
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