0
$\begingroup$

I always thought that it only made sense to take the partial derivative of a function $z=f(x_{1},x_{2},x_{3},...,x_{n})$ with respect to one of its input variables, like ${\partial{f}}/{\partial{x_1}}$. But then I encountered this question:

Compute all first and second partial derivatives, including mixed derivatives, of the following function:$$x^{2}+y^{2}=\sin(xy)$$ Which leads me to think there might be some notion of "implicit partial differentiation" or something thereabouts, but I am quite confused how I should understand this, or what to do. Thanks.

$\endgroup$
0
$\begingroup$

Think of the top most surface of a saddle on a horse. If you think of x and y in the horizontal plane and the elevation of the surface as z and then describe the elevation in terms of the x and y location, then you have a relation.

At some point $(x, y)$ you have defined the z, elevation, and the partials describe how rapidly z changes as you move along x or y from that point.

(This doesn't work very well on a phone.)

$\endgroup$
3
  • $\begingroup$ Yes, I understand that is how partial differentiation works for functions $z(x,y)$ for instance, but how do I understand the partial derivative of a relation where the dependent variable $z$ does not seem to be explicitly isolated? I can tell that constraining $f(x,y)=g(x,y)$ will still amount to some kind of surface in $3$-space, but I'm confused about how precisely to compute the partial derivative of a relation, if that even makes sense. $\endgroup$ – SurfaceIntegral Dec 24 '20 at 17:17
  • $\begingroup$ See hyperphysics.phy-astr.gsu.edu/hbase/deriv.html#c2 for an example. If there were the xy term in the function, its partial would be added. df/dx of x^2 + xy + y^2 = 2x + y $\endgroup$ – Jim Clark Dec 24 '20 at 17:43
  • $\begingroup$ Continuing, See hyperphysics.phy-astr.gsu.edu/hbase/deriv.html#c2 for an example. If there were the xy term in the function, its partial would be added. df/dx of x^2 + xy + y^3 = 2x + y while df/dy = x + 3y^2 $\endgroup$ – Jim Clark Dec 24 '20 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.