Let $s:I\to\mathbb{R}$ a bijective function. Assuming that $h$ is inverse of function $s$ and $c = c(s)$, show that $\frac{dc}{ds}\small{(h(s))} = \frac{dc}{dt}\small{(h(s))}\frac{dh(s)}{ds}$. I don't understand very good the Leibniz notation.
From chain rule we have that $\frac{dc}{ds} = \frac{dc}{dt}\frac{dt}{ds}$. But how can I evaluate this in a point and how transform $dt$ in $dh(s)$? And why the last $\frac{dh(s)}{ds}$ is not evaluated in the point $h(s)$ ?
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MathLearner
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Is $I$ an open interval? – Kenta S Nov 07 '22 at 19:21
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Is a problem derived from differential geometry problem. $c$ is a curve and $s$ is the natural parameter of curve. I don't know if $I$ is open or closed. – MathLearner Nov 07 '22 at 19:24