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The given function is:

\[
G(x, y, z) = \frac{1}{\sqrt{1 - x^2 - y^2 - z^2}}
\]

We are tasked with finding the first partial derivatives of \( G(x, y, z) \) with respect to \(x\), \(y\), and \(z\).

### 1. Partial Derivative with respect to \(x\):

Using the chain rule, we have:

\[
\frac{\partial G}{\partial x} = \frac{\partial}{\partial x} \left( \frac{1}{\sqrt{1 - x^2 - y^2 - z^2}} \right)
\]

First, let's differentiate the outer function \( \frac{1}{u} \) where \( u = \sqrt{1 - x^2 - y^2 - z^2} \):

\[
\frac{d}{du} \left( \frac{1}{u} \right) = -\frac{1}{u^2}
\]

Now, we differentiate the inner function \( u = \sqrt{1 - x^2 - y^2 - z^2} \):

\[
\frac{d}{dx} \left( \sqrt{1 - x^2 - y^2 - z^2} \right) = \frac{1}{2\sqrt{1 - x^2 - y^2 - z^2}} \cdot (-2x)
\]

Thus, the partial derivative with respect to \(x\) becomes:

\[
\frac{\partial G}{\partial x} = -\frac{1}{\left( \sqrt{1 - x^2 - y^2 - z^2} \right)^2} \cdot \left( \frac{x}{\sqrt{1 - x^2 - y^2 - z^2}} \right) = -\frac{x}{(1 - x^2 - y^2 - z^2)^{3/2}}
\]

### 2. Partial Derivative with respect to \(y\):

By symmetry, the partial derivative with respect to \(y\) follows the same steps as for \(x\), but replacing \(x\) with \(y\):

\[
\frac{\partial G}{\partial y} = -\frac{y}{(1 - x^2 - y^2 - z^2)^{3/2}}
\]

### 3. Partial Derivative with respect to \(z\):

Similarly, the partial derivative with respect to \(z\) is:

\[
\frac{\partial G}{\partial z} = -\frac{z}{(1 - x^2 - y^2 - z^2)^{3/2}}
\]

### Summary of Partial Derivatives:
\[
\frac{\partial G}{\partial x} = -\frac{x}{(1 - x^2 - y^2 - z^2)^{3/2}}
\]
\[
\frac{\partial G}{\partial y} = -\frac{y}{(1 - x^2 - y^2 - z^2)^{3/2}}
\]
\[
\frac{\partial G}{\partial z} = -\frac{z}{(1 - x^2 - y^2 - z^2)^{3/2}}
\]

Would you like more details or explanations on any of the steps?

Here are five related questions to further explore:

1. How would you compute the second partial derivatives of \( G(x, y, z) \)?
2. Can you find the gradient vector of \( G(x, y, z) \)?
3. What is the physical interpretation of these partial derivatives in a geometric context?
4. How do partial derivatives relate to the tangent plane of a function?
5. What is the Laplacian of \( G(x, y, z) \) in terms of \(x\), \(y\), and \(z\)?

**Tip:** When differentiating square root expressions, always apply the chain rule, and be careful with the negative signs that arise from differentiating powers and inner functions.

Determine the first partial derivatives of G(x, y, z) = 1 / sqrt(1 - x^2 - y^2 - z^2) with respect to x, y, and z.

This problem explores the first partial derivatives of the function G(x, y, z) = 1 / sqrt(1 - x^2 - y^2 - z^2) with respect to x, y, and z. Learn the step-by-step process using the chain rule and implicit differentiation. Ideal for undergraduate calculus students.

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