Derivative of a function is equal to zero

  • #1
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37

Main Question or Discussion Point

Suppose:

- that I have a function ##g(t)## such that ##g(t) = \frac{dy}{dt} ##;
- that ##y = y(x)## and ##x = x(t)##;
- that I take the derivative of ##g## with respect to ##y##.

One one hand this is ##\frac{dg}{dy} = \frac{dg}{dx}\frac{dx}{dy} = \frac{d^2 y}{dxdt}\frac{dx}{dy}##. On the other hand, if I operate right into ##g = \frac{dy}{dt}## with ##d/dy##, it is ##(d/dy)(dy/dt) = (d/dt)(dy/dy) = 0##. Where is my mistake?
 

Answers and Replies

  • #2
831
37
Sorry, I have edited my post to correct a step
 
  • #3
12,318
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You confused ##y(t)## with ##y(x)##. Write the functions with their variables: ##g=\dfrac{d}{dt}y(t)=g(t)## and ##y=y(x)## and with ##x=x(t)## you have ##y=y(x(t))##.
$$
\dfrac{d}{dy} g = \dfrac{d}{dy} \dfrac{dy}{dt} y(t)= \dfrac{d}{dt}y(t)=\dfrac{d}{dx}y(x) \dfrac{d}{dt}x(t)
$$
What was the part in the middle? Sorry, which was the other approach?
 
  • #4
jim mcnamara
Mentor
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First post is now gone, no question to answer.
 
  • #5
12,318
8,702
Restored for readability.
 

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