Problem 26 - Single Variable Calculus

Question

Differentiate the function. $$ y=x^{2} \log _{2} \sqrt{x^{2}-1} $$

Step-by-Step Solution

Answer

The short answer is: \(\frac{dy}{dx} = \frac{x^3}{(x^2 - 1) \ln{2}} + 2x\log_2{(x^2 - 1)^{\frac{1}{2}}}\).

Step 1: Differentiate \(u(x)\) with respect to \(x\)

Differentiate \(u(x) = x^2\) with respect to \(x\), using the power rule: \[\frac{du}{dx} = 2x\]

Step 2: Differentiate \(v(x)\) with respect to \(x\)

Now, we will differentiate \(v(x) = \log_2 {(x^2-1)^{\frac{1}{2}}}\) with respect to \(x\). To do this, we need to use the chain rule: 1. Differentiate the outer function \(g(z) = \log_2 z\) with respect to \(z\). \[\frac{dg}{dz} = \frac{1}{z\ln{2}}\] 2. Differentiate the inner function \(h(x) = (x^2 - 1)^{\frac{1}{2}}\) with respect to \(x\), using the chain rule: \[\frac{dh}{dx} =\frac{1}{2}(x^2-1)^{-\frac{1}{2}}\cdot 2x\] Now, apply the chain rule: \[\frac{dv}{dx} = \frac{dg}{dz}\cdot \frac{dh}{dx}\] Substitute the derivatives we found above: \[\frac{dv}{dx} = \frac{1}{(x^2 - 1)^{\frac{1}{2}}\ln{2}} \cdot \frac{1}{2}(x^2-1)^{-\frac{1}{2}}\cdot 2x\] Simplify the expression: \[\frac{dv}{dx} = \frac{x}{(x^2 - 1) \ln{2}}\]

Step 3: Apply the product rule to find the derivative of the whole function

Now that we have the derivatives of each expression, we can use the product rule to find the derivative of the whole function. The product rule states that if \(y = u(x)v(x)\), then \(\frac{dy}{dx} = u(x)\frac{dv}{dx} + v(x)\frac{du}{dx}\). So, applying the product rule: \[\frac{dy}{dx}= x^2 \cdot \frac{1}{(x^2-1)\ln{2}} \cdot x + \log_2 {(x^2-1)^{\frac{1}{2}}} \cdot 2x\] Simplifying the expression, we get: \[\frac{dy}{dx} = \frac{x^3}{(x^2 - 1) \ln{2}} + 2x\log_2{(x^2 - 1)^{\frac{1}{2}}}\] So, the derivative of the given function is: \[\boxed{\frac{dy}{dx} = \frac{x^3}{(x^2 - 1) \ln{2}} + 2x\log_2{(x^2 - 1)^{\frac{1}{2}}}}\]

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus used to differentiate functions that are products of two other functions. If you have two functions, say \( u(x) \) and \( v(x) \), and you want to differentiate their product \( y = u(x) v(x) \), the product rule comes into play. It states:\[ \frac{dy}{dx} = u(x) \frac{dv}{dx} + v(x) \frac{du}{dx} \]In simple words, to find the derivative of a product, differentiate each function separately, multiply and then add them together:

• First, multiply the derivative of the first function by the second function.
• Next, multiply the derivative of the second function by the first function.
• Finally, add these two results together.

This rule is crucial when dealing with functions composed of multiple factors, ensuring you account for all aspects of the functions involved.

Chain Rule

The chain rule is an essential differentiation rule used when handling composite functions, which are functions nested within each other. It helps us differentiate a function like \( f(x) = g(h(x)) \), where \( h(x) \) is inside \( g \). The chain rule states:\[ \frac{d}{dx}[g(h(x))] = g'(h(x)) \cdot h'(x) \]Breaking this down makes it simple:

• First, differentiate the outer function as if the inner function is a single variable.
• Next, multiply by the derivative of the inner function.

This rule allows you to peel away each layer of the composite function, making sure the influence of each inner function is considered. In our example, it helped differentiate a logarithmic function applied to a square root.

Logarithmic Differentiation

Logarithmic differentiation is a special technique often used when differentiating functions involving products, quotients, or powers. It simplifies the differentiation process by applying a logarithm to both sides of an equation, utilizing the properties of logarithms to break down multiplication and division into addition and subtraction.Here's the basic idea:

• Take the natural log of both sides of the function: \( y = f(x) \) becomes \( \ln{y} = \ln{f(x)} \).
• Use the properties of logarithms to simplify \( \ln{f(x)} \).
• Differentiate using implicit differentiation, keeping in mind that the derivative of \( \ln{y} \) is \( \frac{1}{y} \frac{dy}{dx} \).
• Solve for \( \frac{dy}{dx} \) to find the derivative of the original function.

This technique is particularly useful in our exercise because it simplifies differentiating complex functions, such as those involving powers and roots, combined with logarithms.

Power Rule

The power rule is one of the most straightforward and frequently used rules in differentiation. It is used to differentiate functions where a variable is raised to a constant power, such as \( x^n \). The power rule can be stated as:\[ \frac{d}{dx} [x^n] = n x^{n-1} \]To apply the power rule:

• Take the exponent \( n \) and use it as a coefficient in front of the variable.
• Subtract one from the exponent to get the new power.

The power rule was applied in our exercise to differentiate \( x^2 \), resulting in the easy-to-follow derivative \( 2x \). It's essential to get comfortable with the power rule, as it appears often in differentiation courses and is foundational for learning more complex calculus concepts.