2

This question comes from the recently-released GRE Math Subject Test Form GR3768. The question is as follows:

Let $u(x,y)$ and $v(x,y)$ be real-valued differentiable functions that are implicitly defined by the equations $x = f(u,v)$ and $y = g(u,v)$, where $f$ and $g$ are real-valued differentiable functions. Which of the following is an expression for $\dfrac{\partial u}{\partial x}$?

The correct answer to this question is $$\frac{\partial u}{\partial x} = \frac{\dfrac{\partial g}{\partial v}}{\dfrac{\partial f}{\partial u} \dfrac{\partial g}{\partial v}-\dfrac{\partial f}{\partial v} \dfrac{\partial g}{\partial u}}$$

if the denominator is different from zero.

I'm stuck on where to start due to the way the functions are named confusing me. That being said, I did attempt to start with replacing the expressions for $x, y$ into the definitions of $u, v$, but I wasn't sure how to deal with the recursive definitions of such functions.

Robert Z
  • 145,942
Stamp
  • 399

2 Answers2

0

I derived this rule in my answer here as a step toward finding the equivalent rule for second derivatives. We can derive the same relationship more succinctly by using the relationship between the Jacobians

$$\begin{pmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{pmatrix} = \begin{pmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{pmatrix}^{-1}$$

The answer follows by equating the $(1,1)$ entry of both matrices

$$\frac{\partial u}{\partial x} = \frac{\frac{\partial y}{\partial v}}{\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}- \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}}$$

and gives all four inverse partial derivative rules at the same time.

Ninad Munshi
  • 34,407
0

$(x, y) = (f(u, v), g(u, v)) =: F(u, v)$ implies $I = DF(u, v)\begin{pmatrix}u_x & u_y \\ v_x & v_y\end{pmatrix}$, so $$\begin{pmatrix}u_x & u_y \\ v_x & v_y\end{pmatrix} = DF(u, v)^{-1}.$$ Now use the formula for the inverse of a 2x2 matrix.

Kakashi
  • 1,381