# Partial Derivative of a Relation?

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.

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.)

• 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. – SurfaceIntegral Dec 24 '20 at 17:17
• 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 – Jim Clark Dec 24 '20 at 17:43
• 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 – Jim Clark Dec 24 '20 at 17:50