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I'm having trouble evaluating the following expression and would appreciate anyone being able to step me through the process: equation

I'm a second-time poster so any suggestions for title/question rephrasing are welcome.

Thanks!

newbie
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  • Your placement of the "$-1$" suggests you want $\frac{\partial}{\partial t} \frac{1}{c(t) \mathrm{e}^{-\rho t}}$. Is that your intent? – Eric Towers Mar 18 '20 at 05:45
  • Also, "solving" is a process applied to equations. You haven't provided an equation. You have provided an expression. Are you having trouble evaluating that expression? – Eric Towers Mar 18 '20 at 05:47
  • Yes to both your corrections. I'll try and edit the post accordingly. – newbie Mar 18 '20 at 05:51

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This is one of the many case where logarithmic differentiation makes life easier$$y(t)=\frac{1}{c(t)\, {e}^{-\rho t}}\implies \log(y(t))=-\log(c(t))+\rho t$$ $$\frac{y'(t)}{y(t)}=-\frac{c'(t)}{c(t)}+\rho$$ Now, use $$y'(t)=y(t)\times \frac{y'(t)}{y(t)}$$ and simplify.

Claude Leibovici
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