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We are tasked with proving the following equation:

\[
\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 = e^{-2s} \left[ \left( \frac{\partial f}{\partial s} \right)^2 + \left( \frac{\partial f}{\partial t} \right)^2 \right]
\]

where \( z = f(x, y) \) and the variables \( x \) and \( y \) are expressed in terms of \( s \) and \( t \) as:

\[
x = e^s \cos t, \quad y = e^s \sin t
\]

### Step-by-step Solution:

#### 1. Chain rule for partial derivatives:
We begin by using the chain rule to express the derivatives of \( f \) with respect to \( x \) and \( y \) in terms of the derivatives with respect to \( s \) and \( t \).

The chain rule for \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) gives:

\[
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial x} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial x}
\]
\[
\frac{\partial f}{\partial y} = \frac{\partial f}{\partial s} \frac{\partial s}{\partial y} + \frac{\partial f}{\partial t} \frac{\partial t}{\partial y}
\]

However, since \( x = e^s \cos t \) and \( y = e^s \sin t \), we directly differentiate these to simplify.

#### 2. Find the partial derivatives of \( x \) and \( y \):
\[
\frac{\partial x}{\partial s} = e^s \cos t, \quad \frac{\partial x}{\partial t} = -e^s \sin t
\]
\[
\frac{\partial y}{\partial s} = e^s \sin t, \quad \frac{\partial y}{\partial t} = e^s \cos t
\]

#### 3. Express derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \):
Using the chain rule, we can now compute \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) in terms of \( s \) and \( t \).

\[
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial s} \cdot \frac{\partial s}{\partial x} + \frac{\partial f}{\partial t} \cdot \frac{\partial t}{\partial x}
\]



Let z = f(x, y) be a function of x and y, where x = e^s * cos(t) and y = e^s * sin(t) are functions of s and t. Prove that (∂f/∂x)^2 + (∂f/∂y)^2 = e^(-2s) * [(∂f/∂s)^2 + (∂f/∂t)^2].

In this advanced calculus problem, learn how to use the chain rule to prove a relationship between the partial derivatives of a function z = f(x, y), where x = e^s * cos(t) and y = e^s * sin(t). The proof involves manipulating partial derivatives and using the chain rule. Suitable for university-level students.

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