Question

Differentiate the following functions: \(\left(y^{2}+1\right) \sqrt{y^{3}-y}\)

Short answer

Question: Find the derivative of the function \((y^2+1) \sqrt{y^3-y}\). Answer: \(h'(y) = 2y\sqrt{y^3 - y} + (y^2+1)\left(\frac{1}{2}(y^3 - y)^{(-\frac{1}{2})} (3y^2 - 1)\right)\).

Step-by-step solution

Identify the functions

Here, we have two functions that are multiplied: \(f(y) = y^2+1\) and \(g(y) = \sqrt{y^3-y}\).

Apply the Product Rule

The product rule states that if we have a function \(h(y) = f(y)g(y)\), then \(h'(y) = f'(y)g(y) + f(y)g'(y)\). In this problem, we will find \(f'(y)\) and \(g'(y)\) first, and then apply the product rule.

Calculate \(f'(y)\)

To find the derivative of \(f(y) = y^2+1\), we use the power rule. The power rule states that if \(f(y) = y^n\), then \(f'(y)=ny^{n-1}\). So, \(f'(y) = 2y^{2-1} = 2y\).

Calculate \(g'(y)\)

To find the derivative of \(g(y) = \sqrt{y^3-y}\), we first rewrite the function in the form \(g(y) = (y^3 - y)^{(\frac{1}{2})}\). Then, we can apply the chain rule which states that \((f \circ g)'(y) = f'(g(y)) \cdot g'(y)\). For our case, let \(u(y) = y^3 - y\). Then, \(g(y) = u(y)^{\frac{1}{2}}\) and we have: - \(u'(y) = \frac{d}{dy}(y^3 - y) = 3y^2 - 1\) - \(g'(y) = \frac{1}{2}u(y)^{(-\frac{1}{2})} \cdot u'(y) = \frac{1}{2}(y^3 - y)^{(-\frac{1}{2})} (3y^2 - 1)\)

Apply the Product Rule

Now that we have found \(f'(y)\) and \(g'(y)\), we can apply the product rule to find the derivative of the given function: \(h'(y) = f'(y)g(y) + f(y)g'(y)\) \(h'(y) = 2y\sqrt{y^3 - y} + (y^2+1)\left(\frac{1}{2}(y^3 - y)^{(-\frac{1}{2})} (3y^2 - 1)\right)\) The derivative of the given function is: \(h'(y) = 2y\sqrt{y^3 - y} + (y^2+1)\left(\frac{1}{2}(y^3 - y)^{(-\frac{1}{2})} (3y^2 - 1)\right)\).

Key concepts

Product Rule

The Product Rule is an essential rule in calculus for differentiating functions that are multiplied together. In cases where we have two functions, such as \( f(y) \) and \( g(y) \), the Product Rule helps to find the derivative of their product accurately. It states that if \( h(y) = f(y) \cdot g(y) \), then the derivative \( h'(y) \) is given by:

• \( h'(y) = f'(y) \cdot g(y) + f(y) \cdot g'(y) \)

This formula ensures that the derivatives of both functions are taken into account. The full expression requires computing \( f'(y) \) and \( g'(y) \) separately. Once these derivatives are calculated, they are plugged into the Product Rule formula to find the overall derivative of the product. This rule is beneficial in a range of situations from basic to more complex functions.

Power Rule

The Power Rule is one of the simplest and most frequently used rules in differentiation. It provides a straightforward method for finding the derivative of a power of \( y \). The rule states that for a function \( f(y) = y^n \), its derivative \( f'(y) \) is given by:

• \( f'(y) = n \cdot y^{n-1} \)

This rule is applicable whenever you have a variable raised to a power. The practical use of this rule simplifies the process of differentiation by reducing the power of \( y \) by one and multiplying by the original power. In our exercise, the expression \( y^2 \) in \( f(y) = y^2 + 1 \) becomes \( f'(y) = 2y \). This example shows how the Power Rule effectively reduces the complexity of taking derivatives.

Chain Rule

The Chain Rule is used when you need to differentiate a composite function. A composite function involves one function inside another, like \( g(y) = (y^3 - y)^{\frac{1}{2}} \). The Chain Rule is crucial for breaking down these functions. It can be described as follows:

• If a function is \( f(g(y)) \), then the derivative \((f\circ g)'(y) = f'(g(y)) \cdot g'(y) \).

The rule implies that we first find the derivative of the outer function \( f \), then multiply it by the derivative of the inner function \( g \). Let's consider the inside function as \( u(y) = y^3 - y \). The derivative of \( u(y) \), denoted \( u'(y) \), is calculated as \( 3y^2 - 1 \). Applying the Chain Rule on \( g(y) = u(y)^{\frac{1}{2}} \), allows us to simplify complex nested functions effectively.

Derivative Calculation

Calculating the derivative of a function involves systematically applying differentiation rules like the Product, Power, and Chain Rules correctly. For our exercise involving the function \( (y^2+1)\sqrt{y^3-y} \), the derivation process involves several steps:

• Firstly, identify the functions' components and the rules that apply.
• Use the Power Rule for components like \( y^2 \).
• Apply the Chain Rule to functions within functions, as shown for \( \sqrt{y^3-y} \).
• Combine all parts using the Product Rule to address the entire expression.

By carefully following these rules, you can compute the derivative. This process helps in understanding how derivative calculation simplifies complex expressions and reveals the behavior of functions with respect to changes in their variables.