Undergrad Derivative of a function is equal to zero

[kent davidge]

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?

 

[kent davidge]

Sorry, I have edited my post to correct a step

 

[fresh_42]

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?

 

[jim mcnamara]

First post is now gone, no question to answer.

 

[fresh_42]

Restored for readability.