Parametric Differentiating

Sometimes we need to write derivatives with respect to $$$x$$$ through differentials of another variable $$$t$$$. In this case expression for derivatives will be more complex.

So, let's calculate differentials with respect to $$$t$$$, in other words $$$x$$$ is not an independent variable.

$$${y}'_{{x}}=\frac{{{d}{y}}}{{{d}{x}}}$$$

$$${y}''_{{{{x}}^{{2}}}}={\left(\frac{{{d}{y}}}{{{d}{x}}}\right)}'_{{x}}=\frac{{{d}{\left(\frac{{{d}{y}}}{{{d}{x}}}\right)}}}{{{d}{x}}}=\frac{{\frac{{{{d}}^{{2}}{y}\cdot{d}{x}-{d}{y}\cdot{{d}}^{{2}}{x}}}{{{d}{{x}}^{{2}}}}}}{{{d}{x}}}=\frac{{{{d}}^{{2}}{y}\cdot{d}{x}-{d}{y}\cdot{{d}}^{{2}}{x}}}{{{d}{{x}}^{{3}}}}$$$ (note that we used quotient rule)

$$${y}'''_{{{{x}}^{{3}}}}={\left(\frac{{{{d}}^{{2}}{y}\cdot{d}{x}-{d}{y}\cdot{{d}}^{{2}}{x}}}{{{d}{{x}}^{{3}}}}\right)}'_{{x}}=\frac{{{d}{\left(\frac{{{{d}}^{{2}}{y}\cdot{d}{x}-{d}{y}\cdot{{d}}^{{2}}{x}}}{{{d}{{x}}^{{3}}}}\right)}}}{{{d}{x}}}=\frac{{\frac{{{d}{{x}}^{{3}}{\left({d}{x}\cdot{{d}}^{{3}}{y}-{{d}}^{{3}}{x}\cdot{d}{y}\right)}-{3}{d}{{x}}^{{2}}\cdot{{d}}^{{2}}{x}{\left({d}{x}\cdot{{d}}^{{2}}{y}-{{d}}^{{2}}{x}\cdot{d}{y}\right)}}}{{{d}{{x}}^{{6}}}}}}{{{d}{x}}}$$$ (note that we used product and quotient rule).

So, $$${y}'''_{{{{x}}^{{3}}}}=\frac{{{d}{x}{\left({d}{x}\cdot{{d}}^{{3}}{y}-{{d}}^{{3}}{x}\cdot{d}{y}\right)}-{3}{{d}}^{{2}}{x}{\left({d}{x}\cdot{{d}}^{{2}}{y}-{{d}}^{{2}}{x}\cdot{d}{y}\right)}}}{{{d}{{x}}^{{5}}}}$$$.

These formulas allows us to perform parametric differentiation.

Suppose that we are given two functions $$${x}={u}{\left({t}\right)}$$$ and $$${y}={v}{\left({t}\right)}$$$, where $$${t}$$$ is independent variable.

When there exist derivatives of $$${x}$$$ and $$${y}$$$ with respect to $$${t}$$$ we can find derivatives of $$${y}$$$ with respect to $$${x}$$$ using above formulas.

Sometimes it is more convenient to express derivatives of $$${y}$$$ with respect to $$${x}$$$ in terms of derivatives (not differentials) of $$${x}$$$ and $$${y}$$$ with respect to $$${t}$$$. We can easily obtain them from differential expressons, by dividing numerator and denominator by $$${d}{t},{d}{{t}}^{{3}},{d}{{t}}^{{5}},\ldots$$$ respectively.

Thus, $$${y}'_{{x}}=\frac{{{d}{y}}}{{{d}{x}}}=\frac{{\frac{{{d}{y}}}{{{d}{t}}}}}{{\frac{{{d}{x}}}{{{d}{t}}}}}=\frac{{{y}'_{{t}}}}{{{x}'_{{t}}}}$$$.

$$${y}''_{{{{x}}^{{2}}}}=\frac{{{{d}}^{{2}}{y}\cdot{d}{x}-{d}{y}\cdot{{d}}^{{2}}{x}}}{{{d}{{x}}^{{3}}}}=\frac{{\frac{{{{d}}^{{2}}{y}}}{{{d}{{t}}^{{2}}}}\cdot\frac{{{d}{x}}}{{{d}{t}}}-\frac{{{d}{y}}}{{{d}{t}}}\cdot\frac{{{{d}}^{{2}}{x}}}{{{d}{{t}}^{{2}}}}}}{{{{\left(\frac{{{d}{x}}}{{{d}{t}}}\right)}}^{{3}}}}=\frac{{{x}'_{{t}}{y}''_{{{{t}}^{{2}}}}-{x}''_{{{{t}}^{{2}}}}{y}'_{{t}}}}{{{{\left({x}'_{{t}}\right)}}^{{3}}}}$$$.

Similarly, we obtain expression for third derivative.

So,

$$${\color{green}{{{y}'_{{x}}=\frac{{{y}'_{{t}}}}{{{x}'_{{t}}}}}}}$$$,

$$${\color{red}{{{y}''_{{{{x}}^{{2}}}}=\frac{{{x}'_{{t}}{y}''_{{{{t}}^{{2}}}}-{x}''_{{{{t}}^{{2}}}}{y}'_{{t}}}}{{{{\left({x}'_{{t}}\right)}}^{{3}}}}}}}$$$,

$$${\color{blue}{{{y}'''_{{{{x}}^{{3}}}}=\frac{{{x}'_{{t}}{\left({x}'_{{t}}{y}'''_{{{{t}}^{{3}}}}-{x}'''_{{{{t}}^{{3}}}}{y}'_{{t}}\right)}-{3}{x}''_{{{{t}}^{{2}}}}{\left({x}'_{{t}}{y}''_{{{{t}}^{{2}}}}-{x}''_{{{{t}}^{{2}}}}{y}'_{{t}}\right)}}}{{{{\left({x}'_{{t}}\right)}}^{{5}}}}}}}$$$.