Chain Rule and Implicit Differentiation

Suppose we have a function of two variables, $f(x,y)=\mathrm{e}^{x^2} \sin y$ .

$\begin{maximasession} f(x,y) := exp(x^2) * sin(y); \maximaoutput* \i1. f(x,y) :=... ...\\ \o1. f\left(x , y\right)\mathbin{:=}\exp x^2 \sin y \\ \end{maximasession}$

We saw earlier what the partial derivatives of were with respect to and . But suppose and are functions of some other variables, for instance,

$\begin{maximasession}[x,y]: [s^2 * t, s * t^2]; \maximaoutput* \i2. [x,y] : [s^2 * t, s * t^2]; \\ \o2. \left[ s^2 t , s t^2 \right] \\ \end{maximasession}$

Then what are the partial derivatives of with respect to and ? Happily for us, Maxima does the chain rule automatically.

$\begin{maximasession} diff(f(x,y), s); \par diff(f(x,y), t); \maximaoutput* \i3.... ...2\right)+2 s t e^{s^4 t^2} \cos \left(s t^2\right) \\ \end{maximasession}$

That was easy. But be warned that the derivative with respect to does not work anymore.

$\begin{maximasession} diff(f(x,y), x); \maximaoutput* \i5. diff(f(x,y), x); \\ ... ...ound M - an error. To debug this try debugmode(true); \\ \end{maximasession}$

We could fix this by using different letters, u and v:

$\begin{maximasession} diff(f(u,v), u); \maximaoutput* \i6. diff(f(u,v), u); \\ \o6. 2 u e^{u^2} \sin v \\ \end{maximasession}$

But that is cheating. A better way to fix it is to kill the relationship between x and s, t.

$\begin{maximasession} kill(x, y); diff(f(x,y), x); \par \maximaoutput* \i7. kill... ...\\ \i8. diff(f(x,y), x); \\ \o8. 2 x e^{x^2} \sin y \\ \end{maximasession}$

Implicit differentiation is relatively easy, given what we have already done. For example, let's find the first partial derivatives of when .

$\begin{maximasession} F: x*y*z + x^2*y^3*z^4 - x - x*z; Fx: diff(F, x); Fy: diff... ...2 y^2 z^4-x z}\over{4 x^2 y^3 z^3+x y-x}} \right] \\ \end{maximasession}$

G. Jay Kerns 2009-12-01