Question

Differentiate the following functions: \(\frac{a-x}{\sqrt{2 a x-x^{2}}}\)

Short answer

Question: Determine the derivative of the function \(h(x) = \frac{a - x}{\sqrt{2ax - x^2}}\). Answer: The derivative of the given function is \(h'(x) = \frac{-2ax\sqrt{2ax - x^2}+x^2\sqrt{2ax - x^2} - (a^2 - 2ax + x^2)(2a - 2x)}{(2ax - x^2)}\).

Step-by-step solution

Identify f(x) and g(x)

In this exercise, we are given the function \(h(x) = \frac{a-x}{\sqrt{2 a x-x^{2}}}\), and we have \(f(x) = a - x\) and \(g(x) = \sqrt{2ax - x^2}\).

Calculate f'(x) and g'(x)

Differentiate \(f(x)\) and \(g(x)\) with respect to \(x\) to get their derivatives: For \(f(x) = a - x\), we have \(f'(x) = -1\). For \(g(x) = \sqrt{2ax - x^2}\), rewrite it as \(g(x) = (2ax - x^2)^{1/2}\). Applying the Chain Rule, we have: \(g'(x) = \frac{1}{2}(2ax - x^2)^{-1/2} \cdot (2a - 2x)\).

Apply the Quotient Rule

Use the Quotient Rule to calculate the derivative of \(h(x)\). Substitute \(f(x), f'(x), g(x),\) and \(g'(x)\) into the Quotient Rule formula: \(h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}\) \(h'(x) = \frac{-1 \cdot \sqrt{2ax - x^2} - (a - x) \cdot \frac{1}{2}(2ax - x^2)^{-1/2} \cdot (2a - 2x)}{(\sqrt{2ax - x^2})^2}\)

Simplify the expression

Simplify the expression of \(h'(x)\): \(h'(x) = \frac{-\sqrt{2ax - x^2} - (a - x) \cdot \frac{1}{\sqrt{2ax - x^2}} \cdot (2a - 2x)}{2ax-x^2}\) \(h'(x) = \frac{-\sqrt{2ax - x^2}(2ax - x^2) - (a - x)(2a - 2x)}{(2ax - x^2)}\) \(h'(x) = \frac{-2ax\sqrt{2ax - x^2}+x^2\sqrt{2ax - x^2} - (a^2 - 2ax + x^2)(2a - 2x)}{(2ax - x^2)}\) Now we have the derivative \(h'(x)\) in its simplified form.

Key concepts

Derivative

Derivatives are central to calculus and represent how a function changes with respect to one of its variables. They give us insight into the behavior and rate of change of the function. In this exercise, we are aiming to find the derivative of the function \[h(x) = \frac{a-x}{\sqrt{2ax-x^{2}}}\].
To do this, we need to understand how to differentiate each part of this function. The derivative of a constant, like \(a\), is zero, whereas the derivative of \(-x\) is \(-1\). These simple rules provide us with the derivatives needed to apply advanced rules like the quotient and chain rules effectively.

Chain Rule

The Chain Rule is a crucial tool in differentiation when dealing with composite functions. When a function is composed of another function, the Chain Rule tells us to differentiate the outer function and then multiply it by the derivative of the inner function.
For our function, we have an inner function \(g(x) = \sqrt{2ax - x^2}\) which can be rewritten as \((2ax - x^2)^{1/2}\). The Chain Rule helps us differentiate this by recognizing:

• The outer function as \((u)^{1/2}\).
• The inner function as \(u = 2ax - x^2\).

Applying the Chain Rule gives us the derivative\[g'(x) = \frac{1}{2}(2ax - x^2)^{-1/2} \cdot (2a - 2x)\].
Thus, the Chain Rule simplifies the process of handling complex expressions.

Simplification

Simplifying expressions in calculus is important for obtaining a concise and manageable result. After applying differentiation rules, like the Quotient and Chain Rules, we often end up with complex expressions. Here, simplification involves combining like terms and reducing fractions to make the derivative clearer and more compact.
In our example, the function \[h'(x) = \frac{-\sqrt{2ax - x^2} - (a - x) \cdot \frac{1}{\sqrt{2ax - x^2}} \cdot (2a - 2x)}{2ax-x^2}\] is further refined to \[h'(x) = \frac{-2ax\sqrt{2ax - x^2}+x^2\sqrt{2ax - x^2} - (a^2 - 2ax + x^2)(2a - 2x)}{(2ax - x^2)}\].
This process of simplification helps us to more easily analyze and interpret the rate of change given by the derivative.

Differentiation Techniques

Differentiation techniques in calculus allow us to find the derivatives of various functions efficiently. In this exercise, several techniques come into play:

• Quotient Rule: This rule is useful when dealing with ratios of functions. The formula \(h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}\) helps us differentiate functions like \(\frac{a-x}{\sqrt{2ax-x^2}}\).
• Basic Differentiation Formulas: Derivatives of constants and simple polynomial terms, like \(-x\), help break down more complex functions.
• Chain Rule: Essential for differentiating nested functions, it simplifies finding derivatives of composite functions.

These techniques combine to tackle complex differentiation tasks by breaking them down into simpler, achievable steps. Each method plays a role in arriving at the final derivative solution. By mastering these techniques, you can efficiently solve diverse problems in calculus.