Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\cos ^{3}\left(\frac{x^{2}}{1-x}\right) $$

Step-by-Step Solution

Answer

The derivative is \( D_{x}y = -3 \cos^2\left(\frac{x^2}{1-x}\right) \sin\left(\frac{x^2}{1-x}\right) \cdot \frac{2x - x^2}{(1-x)^2} \).

Step 1: Identify the Outer Function

The given function is \( y = \cos^3\left(\frac{x^2}{1-x}\right) \). This can be seen as \( u^3 \) where \( u = \cos\left(\frac{x^2}{1-x}\right) \). The outer function is \( u^3 \).

Step 2: Differentiate the Outer Function

Differentiate \( u^3 \) with respect to \( u \) using the power rule: \( \frac{d}{du}(u^3) = 3u^2 \). So the derivative with respect to \( u \) is \( 3\cos^2\left(\frac{x^2}{1-x}\right) \).

Step 3: Identify the Middle Function

The middle function, as previously defined, is \( u = \cos\left(\frac{x^2}{1-x}\right) \). It will be differentiated in the next step.

Step 4: Differentiate the Middle Function

Differentiate \( \cos\left(\frac{x^2}{1-x}\right) \) with respect to \( \frac{x^2}{1-x} \). The derivative of \( \cos(v) \) is \(-\sin(v)\), so the derivative of the middle function is \(-\sin\left(\frac{x^2}{1-x}\right) \).

Step 5: Identify the Inner Function

The innermost function is \( v = \frac{x^2}{1-x} \). This is inside the cosine function and needs to be differentiated.

Step 6: Differentiate the Inner Function

Differentiate \( \frac{x^2}{1-x} \) using the quotient rule. The quotient rule is \( \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2} \). Here, \( f = x^2 \) (with \( f' = 2x \)) and \( g = 1-x \) (with \( g' = -1 \)). So, \( \frac{d}{dx}\left(\frac{x^2}{1-x}\right) = \frac{2x(1-x) - x^2(-1)}{(1-x)^2} = \frac{2x - 2x^2 + x^2}{(1-x)^2} = \frac{2x - x^2}{(1-x)^2} \).

Step 7: Apply the Chain Rule

Apply the chain rule: \( D_{x}y = 3\cos^2\left(\frac{x^2}{1-x}\right) \times (-\sin\left(\frac{x^2}{1-x}\right)) \times \frac{2x - x^2}{(1-x)^2} \). Simplify to get the derivative: \( D_{x}y = -3\cos^2\left(\frac{x^2}{1-x}\right)\sin\left(\frac{x^2}{1-x}\right) \cdot \frac{2x - x^2}{(1-x)^2} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, particularly useful in differentiation. It allows us to differentiate composite functions, which are functions within functions. In our problem, we have a function of the form \( y = \cos^3\left(\frac{x^2}{1-x}\right) \), where the inner function \( v = \frac{x^2}{1-x} \) is nested inside a cosine function, and then raised to the power of three.

To apply the chain rule, follow these steps:

• First, differentiate the outer function, treating the inner function as a single variable. For example, if \( y = u^3 \) where \( u = \cos(\frac{x^2}{1-x}) \), differentiate it to get \( 3u^2 \).
• Next, take the derivative of the middle or nested function, which here is \( \cos(v) \), resulting in \( -\sin(v) \).
• Finally, differentiate the innermost function \( v = \frac{x^2}{1-x} \) using the appropriate rule, which in this case is the quotient rule.
• Multiply all these derivatives together to get the final derivative of the composite function.

In this exercise, using the chain rule significantly simplifies finding the derivative of a complex composition of functions.

Power Rule

The power rule is one of the simplest yet most powerful tools in calculus. It specifically helps us differentiate functions of the form \( u^n \), where \( n \) is a constant exponent. The rule states that if \( y = u^n \), then the derivative \( \frac{d}{du}(u^n) = nu^{n-1} \).

For example, in this problem, our outer function is \( u^3 \). By applying the power rule, the derivative becomes \( 3u^2 \). Here, \( u = \cos(\frac{x^2}{1-x}) \) acts as the base of the power, so it's crucial to apply the power rule initially before moving on with other differentiation rules.

Remember, the power rule is typically the first step when tackling problems involving powers. By quickly reducing the exponent by one and multiplying by the original exponent, it allows us to set the groundwork for further applications of the chain rule when dealing with composite functions.

Quotient Rule

The quotient rule is a method within calculus differentiation used to find the derivative of a function that is a quotient of two other functions. The quotient rule is written as:
\[\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\]
This formula is essential when the function is more complex and contains a fraction. Let’s apply it to our inner function \( v = \frac{x^2}{1-x} \).

• Identify \( f(x) = x^2 \) and \( g(x) = 1-x \).
• Compute the derivatives: \( f'(x) = 2x \) and \( g'(x) = -1 \).
• Substitute these into the quotient rule formula:\[\frac{d}{dx} \left( \frac{x^2}{1-x} \right) = \frac{2x(1-x) - x^2(-1)}{(1-x)^2} \]
• Simplify to get \( \frac{2x - x^2}{(1-x)^2} \).

Remember, the consistency of this rule provides a systematic way to differentiate quotients, even when dealing with nested or composite functions. Understanding how to effectively utilize it is key when encountering any differential quotient in calculus.