Question

Differentiate the following funetions: \(y=\frac{1}{\sqrt{a^{2}-x^{2}}}\)

Short answer

Answer: The derivative of the function is \(y' = \frac{x}{(a^2 - x^2)^{\frac{3}{2}}}\).

Step-by-step solution

Identify the Quotient Rule

Recall that the Quotient Rule states that the derivative of a quotient of functions is given by: \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\), where \(u\) is the numerator function, \(v\) is the denominator function, and \(u'\) and \(v'\) are their respective derivatives. In our given function, \(y = \frac{1}{\sqrt{a^2 - x^2}}\), let's identify \(u\) and \(v\): \(u = 1\) (the numerator) \(v = \sqrt{a^2 - x^2}\) (the denominator)

Differentiate \(u\) and \(v\)

First, differentiate the numerator function \(u\) with respect to \(x\). Since \(u\) is a constant function (1), its derivative is 0: \(u' = 0\) Next, differentiate the denominator function \(v\) with respect to \(x\). We will use the Chain Rule because the function is of the form \(v = \sqrt{a^2 - x^2} = (a^2-x^2)^{\frac{1}{2}}\). The Chain Rule states that: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\) In this case, let's identify \(f(g(x))\) and \(g(x)\): \(f(g(x)) = (a^2 - x^2)^{\frac{1}{2}}\) \(g(x) = a^2 - x^2\) Now, differentiate \(g(x)\): \(g'(x) = -2x\) To differentiate \(f(g(x))\), apply the Chain Rule: \(v' = \frac{d}{dx}(a^2 - x^2)^{\frac{1}{2}} = \frac{1}{2}(a^2 - x^2)^{-\frac{1}{2}} \cdot (-2x)\)

Apply the Quotient Rule

Using the Quotient Rule with our identified functions and their derivatives: \(y' = \frac{u'v - uv'}{v^2} = \frac{(0)\cdot (\sqrt{a^2 - x^2}) - (1)\cdot \frac{1}{2}(a^2 - x^2)^{-\frac{1}{2}}(-2x)}{(\sqrt{a^2 - x^2})^2}\)

Simplify the expression

Simplify the expression for \(y'\): \(y' = \frac{-(-x)(a^2 - x^2)^{-\frac{1}{2}}}{a^2 - x^2} = \frac{x}{(a^2 - x^2)^{\frac{3}{2}}}\) The derivative of the given function is: \(y' = \frac{x}{(a^2 - x^2)^{\frac{3}{2}}}\)

Key concepts

Quotient Rule

The Quotient Rule is a fundamental tool in calculus for finding the derivative of a quotient of two functions. When you have a function of the form \( y = \frac{u}{v} \), the rule states that the derivative \( y' \) is given by:

• \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \)

This formula helps in breaking down the process into manageable steps by focusing on the numerator \( u \) and the denominator \( v \), and their respective derivatives \( u' \) and \( v' \). For the exercise \( y = \frac{1}{\sqrt{a^2 - x^2}} \), we identified that:

• \( u = 1 \) (a constant, thus \( u' = 0 \))
• \( v = \sqrt{a^2 - x^2} \) (requires further differentiation)

The Quotient Rule becomes particularly useful when differentiating complex rational functions. Keep in mind that it elegantly handles cases where simply differentiating the functions separately becomes cumbersome.

Chain Rule

The Chain Rule is particularly useful when dealing with compositions of functions, especially when functions are nested within one another. The rule is often stated as:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

This means the derivative of a complex composition \( f(g(x)) \) is the product of the derivative of the outer function \( f \) evaluated at \( g(x) \) and the derivative of the inner function \( g(x) \).
For the given exercise, the function \( v = \sqrt{a^2 - x^2} = (a^2 - x^2)^{\frac{1}{2}} \) was differentiated using the Chain Rule:

• Identify \( f(g(x)) = (a^2 - x^2)^{\frac{1}{2}} \)
• \( g(x) = a^2 - x^2 \)
The intermediate result is \( g'(x) = -2x \), and after applying the Chain Rule, we compute:
• \( v' = \frac{1}{2}(a^2 - x^2)^{-\frac{1}{2}} \times (-2x) \)
Understanding the Chain Rule fully allows one to tackle a wide range of function differentiation problems with confidence.

Calculus

Calculus is a branch of mathematics that provides tools for understanding changes, often used to find rates of change and areas under curves. Differentiation, a primary concept in calculus, involves finding the derivative, which represents the rate at which a function changes at any given point.
In the exercise provided, we differentiated the function \( y = \frac{1}{\sqrt{a^2 - x^2}} \). Differentiation required both the Quotient Rule and the Chain Rule—typical strategies in calculus.

• The Quotient Rule helped with handling the rational part.
• The Chain Rule was used for the nested square root function.

Together, these rules allowed us to find the derivative \( y' = \frac{x}{(a^2 - x^2)^{\frac{3}{2}}} \). Such problems illustrate the power of calculus in simplifying and solving complex mathematical challenges.