# 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,

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

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

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

Example 1. Find $\frac{{{d}{y}}}{{{d}{x}}}$ and $\frac{{{{d}}^{{2}}{y}}}{{{d}{{x}}^{{2}}}}$ if ${y}{\left({t}\right)}={3}{{t}}^{{3}}+{5}{t}$ and ${x}{\left({t}\right)}={2}{{t}}^{{4}}+{7}{{t}}^{{2}}$.

We have that ${y}'_{{t}}={9}{{t}}^{{2}}+{5}$ and ${y}''_{{{{t}}^{{2}}}}={18}{t}$. Now, ${x}'_{{t}}={8}{{t}}^{{3}}+{14}{t}$ and ${x}''_{{{{t}}^{{2}}}}={24}{{t}}^{{2}}+{14}$.

So,

${y}'_{{x}}=\frac{{{d}{y}}}{{{d}{x}}}=\frac{{{9}{{t}}^{{2}}+{5}}}{{{8}{{t}}^{{3}}+{14}{t}}}$.

${y}''_{{{{x}}^{{2}}}}=\frac{{{{d}}^{{2}}{y}}}{{{d}{{x}}^{{2}}}}=\frac{{{\left({8}{{t}}^{{3}}+{14}{t}\right)}{18}{t}-{\left({24}{{t}}^{{2}}+{14}\right)}{\left({9}{{t}}^{{2}}+{5}\right)}}}{{{{\left({8}{{t}}^{{3}}+{14}{t}\right)}}^{{3}}}}$.

Note, that formulas for second derivative and third are fairly complex.

So, it is better to derive these formulas while solving the problem.

Example 2. Find $\frac{{{{d}}^{{2}}{y}}}{{{d}{{x}}^{{2}}}}$ if ${y}{\left({t}\right)}={{t}}^{{2}}+{1}$ and ${x}{\left({t}\right)}={{t}}^{{2}}+{t}$.

Let's first find first derivative:

${y}'_{{x}}=\frac{{{y}'_{{t}}}}{{{x}'_{{t}}}}=\frac{{{2}{t}}}{{{2}{t}+{1}}}$.

We can find second derivative using above formula, but we will take advantage of first derivative.

$\frac{{{{d}}^{{2}}{y}}}{{{d}{{x}}^{{2}}}}=\frac{{d}}{{{d}{x}}}\frac{{{d}{y}}}{{{d}{x}}}=\frac{{\frac{{d}}{{{d}{t}}}{\left(\frac{{{d}{y}}}{{{d}{x}}}\right)}}}{{\frac{{{d}{x}}}{{{d}{t}}}}}=\frac{{\frac{{d}}{{{d}{t}}}{\left(\frac{{{2}{t}}}{{{2}{t}+{1}}}\right)}}}{{\frac{{{d}{x}}}{{{d}{t}}}}}=\frac{{\frac{{{\left({2}{t}\right)}'{\left({2}{t}+{1}\right)}-{2}{t}{\left({2}{t}+{1}\right)}'}}{{{\left({2}{t}+{1}\right)}}^{{2}}}}}{{{2}{t}+{1}}}=\frac{{{4}}}{{{\left({2}{t}+{1}\right)}}^{{3}}}$.