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Suppose I have a function $x(t)$ such that $\frac{d}{dt}x=x(t)y(t)$. Can I affirm hat $$\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}=\frac{1}{x(t)y(t)}$$ I am sorry but I am a bit confused about this argument

QED
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user268193
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  • The function $x$ has to be invertible and, both the function and its inverse has to be differentiable, and also the derivative can not be $0$. – QED Sep 24 '20 at 07:57
  • Assuming the hypotheses specified by QED are satisfied, you can start with $x(t)$ and its inverse function $t(x)$. By definition you have $t(x(\tau)) = \tau$. Now differentiate both sides with respect to $\tau$. – dfnu Sep 24 '20 at 08:52

2 Answers2

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When both are defined you have $$\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}=\frac{1}{x(t)y(t)}.$$

Alessio K
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You are combining two issues in one.

  1. for a differentiable function with nonzero derivative, it is true that $$\dfrac{dx}{dt}=\dfrac1{\dfrac{dt}{dx}}$$ by differential calculus.

  2. if $f(t)=x(t)\,y(t)\ne0$, then $\dfrac1{f(t)}=\dfrac1{x(t)\,y(t)}.$ This is elementary arithmetic.

  • Thank you for your answer. My question was about the first issue. Therefore I can prove that using the limit of the difference quotient? – user268193 Sep 24 '20 at 08:24
  • @user268193: a proof can be based on the identity $$\frac {x(t+h)-x(t)}h=\frac k{t(x+k)-t(x)}$$ where $x(t+h)=x(t)+k$. –  Sep 24 '20 at 08:37