To find the differentiation of \( x(1-y^2)^{1/2} \) with respect to \( x \), we will use the product rule and the chain rule. 

Let \( f(x) = x(1 - y^2)^{1/2} \).

The product rule states that if you have a function that is the product of two functions, say \( u(x) \cdot v(x) \), then the derivative is given by:

\[ (u \cdot v)' = u' \cdot v + u \cdot v' \]

In this case, let:

\[ u(x) = x \]
\[ v(x) = (1 - y^2)^{1/2} \]

First, differentiate \( u(x) \) with respect to \( x \):

\[ u'(x) = \frac{d}{dx}(x) = 1 \]

Next, differentiate \( v(x) \) with respect to \( x \) using the chain rule. Notice that \( v(x) \) does not explicitly depend on \( x \) but on \( y \), which might be a function of \( x \). However, for simplicity, we will assume \( y \) is constant with respect to \( x \):

\[ v'(x) = \frac{d}{dx}\left((1 - y^2)^{1/2}\right) = 0 \]

Hence, applying the product rule:

\[ \frac{d}{dx} \left( x(1 - y^2)^{1/2} \right) = x \cdot 0 + 1 \cdot (1 - y^2)^{1/2} = (1 - y^2)^{1/2} \]

Thus, the derivative of \( x(1 - y^2)^{1/2} \) with respect to \( x \) is \( (1 - y^2)^{1/2} \).

Do you want more details or have any questions?

Here are five questions you might be interested in:
1. How do you apply the chain rule in differentiation?
2. Can you explain the product rule with another example?
3. What is implicit differentiation and when is it used?
4. How do you differentiate trigonometric functions?
5. Can you solve an example involving differentiation of exponential functions?

**Tip:** When differentiating products of functions, always identify and clearly write out the two functions and their derivatives before applying the product rule to avoid confusion.

Learn how to differentiate x times the square root of 1 minus y squared with respect to x. Step-by-step solution using the product rule and chain rule. Detailed explanation suitable for college-level calculus students.

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