Problem 44 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. $$ f(x)=\ln (\cos (1-x)) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \tan(1-x) \).

Step 1: Identify the Function to Differentiate

The function given is \( f(x) = \ln(\cos(1-x)) \). Here, we have a composite function where the natural logarithm function is applied to \( \cos(1-x) \). We need to apply chain rule to differentiate it with respect to \( x \).

Step 2: Apply the Chain Rule

To differentiate \( f(x) = \ln(\cos(1-x)) \), we will use the chain rule. The chain rule states that if you have \( f(g(x)) \), then \( f'(g(x)) \cdot g'(x) \). Here, \( f(u) = \ln(u) \) and \( u = \cos(1-x) \).

Step 3: Differentiate the Outer Function

Differentiate the outer function \( \ln(u) \) with respect to \( u \). The derivative of \( \ln(u) \) is \( \frac{1}{u} \). So, for \( f(u) = \ln(\cos(1-x)) \), the derivative is \( \frac{1}{\cos(1-x)} \).

Step 4: Differentiate the Inner Function

Now differentiate the inner function \( u = \cos(1-x) \) with respect to \( x \). The derivative of \( \cos(v) \) is \( -\sin(v) \), and for \( v = 1-x \), the derivative is \( -1 \). Therefore, \( \frac{d}{dx}[\cos(1-x)] = \sin(1-x) \) because of an additional \(-\sin\) term from differentiating the inside \( 1-x \).

Step 5: Apply the Chain Rule to Obtain the Derivative

Now, apply the chain rule result from Steps 3 and 4: \( f'(x) = \frac{1}{\cos(1-x)} \cdot \sin(1-x) \). This simplifies to \( \frac{\sin(1-x)}{\cos(1-x)} \), which is equal to \( \tan(1-x) \) using the identity \( \tan(v) = \frac{\sin(v)}{\cos(v)} \).

Key Concepts

Chain Rule

When you encounter a composite function, like \(f(x) = \ln(\cos(1-x))\), using the chain rule is essential to find its derivative. The chain rule helps differentiate functions nested inside one another.
It states that if you have a composite function \(f(g(x))\), the derivative is found by multiplying the derivative of the outer function by the derivative of the inner function. This can be written as \(f'(g(x)) \cdot g'(x)\). For our problem, \(f(u) = \ln(u)\) and \(u = \cos(1-x)\). The outer function \(f(u)\) differentiates to \(\frac{1}{u}\), and the inner function \(u\) differentiates to \(\sin(1-x)\) thanks to chain rule applications.
So, when you multiply them, you get the derivative of the original compound function.

• Differentiate outer: \(\frac{d}{du}[\ln(u)] = \frac{1}{u}\)
• Differentiate inner: \(\frac{d}{dx}[\cos(1-x)] = \sin(1-x)\)
• Apply chain rule: \(\frac{1}{\cos(1-x)} \cdot \sin(1-x) = \tan(1-x)\)

Composite Functions

A composite function is formed when one function is applied to the result of another. The notation \(f(g(x))\) shows that \(g(x)\) is inside \(f(x)\). In our problem, \(g(x) = \cos(1-x)\) and \(f(u) = \ln(u)\).
Understanding composite functions is crucial because you need this knowledge to apply the chain rule effectively.
Without analyzing the layers or the ordering of these functions, stepping into differentiation would become tedious.

• Outer function: \(\ln\)
• Inner function: \((\cos, (1-x))\)

When you spot a composite function, focus on identifying the core structure of outer and inner functions. This will guide you through correctly handling differentiation using the chain rule.

Trigonometric Derivatives

Trigonometric functions such as sine and cosine have distinct derivatives that you use often. Knowing these derivatives helps you break down parts of a composite function quickly.
For the \(\cos\) function, the derivative is \(-\sin\).
Meanwhile, the derivative of \(\sin\) is \(\cos\). These derivatives are critical when differentiating anything that includes trigonometric parts.

• Derivative of \(\sin(x): \cos(x)\)
• Derivative of \(\cos(x): -\sin(x)\)

In our problem, applying the derivative correctly to \(\cos(1-x)\) results in multiplying by \(\sin(1-x)\), reflecting that negative transformation of the \(\cos\) function from its derivative. Understanding these rules is one of the first steps in mastering calculus.