Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the below system: $$u^3+v^3+x^5+4y^5=10\\\ u^2+v^2+x^9+y^8=11$$ I was asked by my students to find $u_x,v_y$ where $x,y$ are our independent variables. Please give me some hints. Thank you!

share|cite|improve this question
Hint: Implicit differentiation. – Daryl Jan 18 '13 at 6:17
Since when students give exercises to teachers? – Fabian Jan 18 '13 at 6:30
up vote 1 down vote accepted

The two equations $$\eqalign{F(x,y, u,v)&:=u^3+v^3+x^5+4y^5-10=0\ ,\cr G(x,y,u,v)&:= u^2+v^2+x^9+y^8-11=0\cr}$$ define a two-dimensional surface $S\subset{\mathbb R}^4$. This surface may have self-intersections, bubbles, cusps, and many other kinds of singularities. Therefore you cannot expect that there are "global" functions $$(x,y)\mapsto u(x,y)\ ,\qquad (x,y)\mapsto v(x,y)\qquad(1)$$ such that $S$ appears as graph of the function pair $(u,v)$ in the form $$S=\bigl\{(x,y,u,v)\ \bigm|\ (x,y)\in{\mathbb R}^2,\ u=u(x,y),\ v=v(x,y)\bigr\}\ .$$ But locally such a representation is possible, even if you are not able to solve the given equations for $u$ and $v$ algebraically. This means you are not able to write down $u(x,y)$, $v(x,y)$ explicitly; but such functions are guaranteed to exist; for details see below.

Assume that you have found, e.g. numerically, a point ${\bf p}=(x_0,y_0,u_0,v_0)\in S$. Then under a certain technical assumption there is a $2\times2$-dimensional box $B=Z\times W$ with center ${\bf p}$ such that the part of $S$ lying in this box can be described in the form $$S\cap B=\bigl\{(x,y,u,v)\ \bigm|\ (x,y)\in Z,\ u=u(x,y),\ v=v(x,y)\bigr\}$$ with certain differentiable functions $(1)$ defined in $Z$. It follows that for $(x,y)\in Z$ one has $$F\bigl(x,y,u(x,y),v(x,y)\bigr)=0\ ,\qquad G\bigl(x,y,u(x,y),v(x,y)\bigr)=0\ .$$ Differentiating with respect to $x$ and to $y$ we get four more identities $$\eqalign{F_x + F_u u_x+ F_v v_x&\equiv0\ , \qquad F_y + F_u u_y+ F_v v_y\equiv0\ ,\cr G_x + G_u u_x+ G_v v_x&\equiv0\ ,\qquad G_y + G_u u_y+ G_v v_y\equiv0\ .\cr}\qquad(2)$$ In particular they hold at $(x_0,y_0)$, for which $\bigl(x_0,y_0,u(x_0,y_0),v(x_0,y_0)\bigr)$ is nothing else but the point ${\bf p}=(x_0,y_0,u_0,v_0)$ we started with. The two lefthand equations $(2)$ can be solved for $u_x(x_0,y_0)$ and $v_x(y_0,y_0)$, the two righthand equations $(2)$ for $u_y(x_0,y_0)$, $v_y(x_0,y_0)$ $-$ under the crucial assumption that the determinant of the two equations, which is the same in both cases, is $\ne0$. This is the "technical assumption" referred to above. It reads $$F_u G_v -F_v G_u \bigr|_{\bf p}\ne 0\ .$$

share|cite|improve this answer

HINT $$u^3+v^3+x^5+4y^5=10\\\ u^2+v^2+x^9+y^8=11$$ considering $y$ as constant $$3u^2u_x+3v^2v_x+5x^4=0\\\ 2uu_x+2vv_x+9x^8=0$$ considering $x$ as constant $$3u^2u_y+3v^2v_y+20y^4=0\\\ 2uu_y+2vv_y+8y^7=0$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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