Problem 23 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \log _{2}\left(\arccos \left(x^{2}\right)\right) $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{2x}{\arccos(x^2) \ln(2) \sqrt{1 - x^4}}\).

Step 1: Identify the Structure

The given function is \(y = \log_{2}( ext{arccos}(x^2))\). This function is a composite function involving a logarithm base 2 and an arccosine function.

Step 2: Use the Chain Rule

To differentiate a composite function, we'll apply the chain rule. We set \(u = \arccos(x^2)\) with respect to \(x\), and then we differentiate the function \(y = \log_{2}(u)\) with respect to \(u\). The chain rule states: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

Step 3: Differentiate the Outer Function

The outer function is \(y = \log_{2}(u)\). The derivative of \(\log_{b}(u)\) is given by \(\frac{1}{u \ln(b)}\). Thus, \(\frac{dy}{du} = \frac{1}{u \ln(2)}\).

Step 4: Differentiate the Inner Function

The inner function is \(u = \arccos(x^2)\). The derivative of \(\arccos(v)\) with respect to \(v\) is \(-\frac{1}{\sqrt{1 - v^2}}\). Using the chain rule again for \(v = x^2\), \(\frac{dv}{dx} = 2x\). Thus, \(\frac{du}{dx} = -\frac{2x}{\sqrt{1 - (x^2)^2}} = -\frac{2x}{\sqrt{1 - x^4}}\).

Step 5: Apply the Chain Rule

Combining the results from the previous steps, we have: \(\frac{dy}{dx} = \frac{1}{u \ln(2)} \cdot \frac{du}{dx}\). Substituting \(u = \arccos(x^2)\) and \(\frac{du}{dx}\), we get: \[\frac{dy}{dx} = \frac{1}{\arccos(x^2) \ln(2)} \cdot \left(-\frac{2x}{\sqrt{1 - x^4}}\right) = -\frac{2x}{\arccos(x^2) \ln(2) \sqrt{1 - x^4}}\].

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions, which are expressions involving multiple functions applied one within another. Imagine you have a function, say \( y = f(g(x)) \), where \( g(x) \) is nested inside \( f(x) \). To differentiate \( y \) with respect to \( x \), the chain rule tells us to take the derivative of the outer function \( f \) with respect to \( g \, (f'(g(x))) \), and then multiply it by the derivative of the inner function \( g \) with respect to \( x \, (g'(x)) \). Thus, the formula for the chain rule is:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

By applying the chain rule, you can differentiate complex expressions step by step. In our exercise, this technique is critical because the function involves nested expressions such as \( \arccos(x^2) \) inside the logarithm base 2 function. Each part is treated sequentially to find the final derivative.

Composite Functions

Composite functions involve the application of one function inside another, creating a layered effect. In calculus, solving problems with composite functions means we must unravel these layers. Consider the expression \( y = \, \log_{2}(\arccos(x^2)) \). This is a composite function:

• The innermost function is \( x^2 \).
• The next function applied is \( \arccos(x^2) \).
• The outermost function is \( \log_{2}\) of what comes from the arccosine function.

When differentiating a composite function, we tackle one layer of the function after another, moving from the outside inward or vice versa depending on the approach. Using the chain rule allows us to deal with each part effectively, helping in obtaining the derivative of such composite expressions efficiently.

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique particularly useful for differentiating functions that involve variables in the exponent or under other transcendental functions like logarithms. This technique involves taking the natural logarithm (\( \ln \)) of a function and differentiating using properties of logarithms.For example, if you have a function \( y = a^x \), where \( a \) is a constant, differentiating directly might be complex, but by taking the logarithm on both sides:

• \( \ln(y) = x \ln(a) \)
• And differentiating gives: \( \frac{1}{y} \cdot \frac{dy}{dx} = \ln(a) \), allowing for a simpler solution path.

In our specific scenario with \( y = \log_{2}(\arccos(x^2)) \), logarithmic properties assist in understanding how the base of the logarithm influences the derivative with respect to \( x \). By leveraging logarithmic differentiation, we can simplify and solve complex derivatives that involve logarithms.

Arccosine Function

The arccosine function, denoted as \( \arccos(x) \), is the inverse of the cosine function on a restricted domain. Differentiating this function requires extra care due to its inverse nature and the need to respect its domain: \( 0 \leq x \leq 1 \), since the range of the function is related to angles.The derivative of \( \arccos(x) \) is given by:

• \( \frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1 - x^2}} \)

This arises because the chain rule and inverse trigonometric identities are used. In the equation from our exercise, \( x^2 \) is the argument of \( \arccos \), meaning we must additionally apply the chain rule to incorporate the derivative of \( x^2 \) as in \( \arccos(x^2) \). This distinction is critical for accurately determining the rate of change of such functions.