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in an introductory text on calculus for economists I found the following $$ y=f(\bar{h}-t) $$ then, differentiating $y$ in respect to $t$ $$ \frac{dy}{dt}=f'(\bar{h}-t)\frac{d}{dt}(\bar{h}-t) $$ then simplifying $$ \frac{dy}{dt}=-f'(\bar{h}-t) $$ While I think I understood the chain rule-based differentiation (derivative of outer function times inner function times derivative of inner function), I struggle to work out the simplification. Can please anyone be so patient to help with the simplification? Thank you

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  • $\begingroup$ This is correct. There is not a simplification, it just computed the derivative and treated $\overline y$ as a constant nothing more. WHich i dont know can be true :). $\endgroup$ – Brethlosze Jun 4 '17 at 15:54
  • $\begingroup$ @hyprfrco You mean $\overline {h} $ $\endgroup$ – hamam_Abdallah Jun 4 '17 at 16:07
  • $\begingroup$ yes!.. my mistake $\endgroup$ – Brethlosze Jun 4 '17 at 16:18
  • $\begingroup$ Thank you very much. Now it makes sense, I got confused because the author said he was doing simplification. $\endgroup$ – Nicola Pensiero Jun 4 '17 at 18:06
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As you said, first you take the outer derivative, resulting in the prime notation. Then you take the inner derivative, which is equal to -1. Multiplying then together gives the result. It isn't simplified, just a slightly odd notation I think. Obviously, all this is assuming h is not a function of t.

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