Question

Finding a Derivative In Exercises \(7-34,\) find the derivative of the function. $$ y=x \sqrt{1-x^{2}} $$

Short answer

The derivative of the function \(y = x \sqrt{1 - x^{2}}\) is \(y' = \sqrt{1 - x^{2}} - x^{2}/ \sqrt{1 - x^{2}}\)

Step-by-step solution

Identify the Component Functions

First, identify the two functions that are being multiplied together in this case: \(u(x) = x\) and \(v(x) = \sqrt{1 - x^{2}}\) or written in power form \(v(x) = (1 - x^{2})^{0.5}\).

Compute the Derivatives

Next, find the derivatives of \(u(x)\) and \(v(x)\), denoted as \(u'(x)\) and \(v'(x)\) respectively. The derivative of \(u(x)\) is \(u'(x) = 1 \). The derivative of \(v(x)\) requires the use of the chain rule, \( (f(g(x)))' = f'(g(x))g'(x)\), where f(g(x)) is \((1 - x^{2})^{0.5}\). We treat \(f(x) = x^{0.5}\) and \(g(x) = 1 - x^{2}\), which gives us \(f'(x) = 0.5x^{-0.5}\) and \(g'(x) = -2x\). Applying chain rule, we get: \( v'(x) = f'(g(x))g'(x) = 0.5(1 - x^{2})^{-0.5} * -2x = -x (1 - x^{2})^{-0.5}\).

Apply the Product Rule

Finally, we can apply the product rule of derivatives: \((uv)' = u'v + uv'\). Plugging in we get: \(y' = u'v + uv' = 1 * \sqrt{1 - x^{2}} + x * (-x(1 - x^{2})^{-0.5}) = \sqrt{1 - x^{2}} - x^{2}(1 - x^{2})^{-0.5}\). This simplifies to \(y' = \sqrt{1 - x^{2}} - x^{2}/ \sqrt{1 - x^{2}} \)

Key concepts

product rule

When dealing with derivatives of functions that are products of two functions, the **product rule** is the go-to technique. This rule is great because it lets us find derivatives without having to multiply out the functions, which can get complicated. The key formula is

• \((uv)' = u'v + uv'\)

Here, \(u\) and \(v\) are two separate functions of \(x\), and \(u'\) and \(v'\) are their respective derivatives. You simply multiply the derivative of the first function by the second function, then add it to the first function times the derivative of the second function.
Consider the function \(y = x \sqrt{1 - x^2}\). The two functions are \(u(x) = x\) and \(v(x) = \sqrt{1 - x^2}\). Applying the product rule simplifies the process of differentiation, allowing us to find that

• \(y' = \sqrt{1 - x^{2}} - x^{2}/ \sqrt{1 - x^{2}} \)

chain rule

The **chain rule** is your best friend when you're dealing with compositions of functions. This rule helps you find the derivative of a function within another function, using the formula:

• \((f(g(x)))' = f'(g(x))g'(x)\)

In this expression, \(f(x)\) represents the outer function and \(g(x)\) is the inner function. You differentiate the outer function first, keeping the inner function unchanged, and then multiply by the derivative of the inner function.
For our example, \(v(x) = (1-x^2)^{0.5}\), we treated the outer function as \(f(x) = x^{0.5}\) and the inner function as \(g(x) = 1 - x^{2}\). Differentiating these gives us \(f'(x) = 0.5x^{-0.5}\) and \(g'(x) = -2x\). Thus, applying the chain rule results in

• \(v'(x) = -x (1 - x^{2})^{-0.5}\)

power rule

The **power rule** is perhaps the most straightforward differentiation rule out there. It's particularly handy when you're dealing with polynomials and allows you to quickly find derivatives of power functions.

• If \(y = x^n\), then the derivative \(y' = nx^{n-1}\)

This means you multiply the function by its power, then reduce the power by one. In our original problem, we needed to express \(v(x) = \sqrt{1 - x^2}\) in a form compatible with other rules, which we wrote as \( (1 - x^2)^{0.5}\). Although using the power rule directly here wasn’t necessary, expressing functions in power form makes them more accessible for other rules like the chain rule.