Problem 57 - Essential Calculus Early Transcendentals

Question

\(51-60=\) Use logarithmic differentiation or an alternative method to find the derivative of the function. $$ y=(\cos x)^{x} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = (\cos x)^x (\ln(\cos x) - x \tan x)\)

Step 1: Take the Natural Logarithm

Start by taking the natural logarithm of both sides of the equation to help differentiate more easily. Applying the logarithm allows us to bring the exponent down as a coefficient.\[\ln y = \ln((\cos x)^x) = x \ln(\cos x)\]

Step 2: Differentiate Using Implicit Differentiation

Differentiate both sides with respect to \(x\). Use implicit differentiation and the product rule on the right-hand side.The left side becomes \(\frac{1}{y} \cdot \frac{dy}{dx}\). The right side becomes:\[\frac{d}{dx}[x \ln(\cos x)] = \ln(\cos x) + x \cdot \frac{d}{dx}[\ln(\cos x)]\]For \(\frac{d}{dx}[\ln(\cos x)]\), apply the chain rule:\[\frac{d}{dx}[\ln(\cos x)] = - \tan x\]Combine these:\[\ln(\cos x) - x \tan x\]

Step 3: Isolate \(\frac{dy}{dx}\)

From the differentiated equation, solve for \(\frac{dy}{dx}\) by multiplying both sides by \(y\):\[\frac{1}{y} \cdot \frac{dy}{dx} = \ln(\cos x) - x \tan x\]Multiply through by \(y\):\[\frac{dy}{dx} = y(\ln(\cos x) - x \tan x)\]Now replace \(y\) with \((\cos x)^x\) as per the original function:\[\frac{dy}{dx} = (\cos x)^x (\ln(\cos x) - x \tan x)\]

Step 4: Final Expression for the Derivative

The final expression for the derivative of the function \(y = (\cos x)^x\) is:\[\frac{dy}{dx} = (\cos x)^x \left(\ln(\cos x) - x \tan x\right)\]

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function that is not explicitly solved for one variable in terms of another. Often, you'll encounter equations where a dependent variable, like \( y \), is intertwined with an independent variable \( x \) in such a way that solving for \( y \) first is not straightforward.

Instead of isolating \( y \), we differentiate both sides of an equation with respect to \( x \) directly. This involves treating \( y \) as an implicit function of \( x \). When taking the derivative of \( y \) with respect to \( x \), you apply the derivative symbol \( \frac{dy}{dx} \) to \( y \) as if it were a function of \( x \).

• Differentiate both sides of an equation relative to \( x \).
• Remember to multiply each derivative of \( y \) by \( \frac{dy}{dx} \).
• Solve the resulting equation for \( \frac{dy}{dx} \).

This method proved handy in differentiating \( \ln y = x \ln(\cos x) \) in our original step-by-step solution, allowing us to manage the derivative of \( y \) indirectly.

Chain Rule

The chain rule is a fundamental differentiation rule in calculus that helps us differentiate composite functions, where one function nests inside another. The rule articulates that to differentiate a composite function \( f(g(x)) \), you take the derivative of the outer function evaluated at the inner function, \( f'(g(x)) \), and multiply it by the derivative of the inner function, \( g'(x) \).

In the original exercise, after transforming \( y = (\cos x)^x \) using logarithms to \( \ln y = x \ln(\cos x) \), we applied the chain rule to take the derivative of \( \ln(\cos x) \), which resulted in effectively handling the nested functions.

• Identify the outer and inner functions. For example, in \( \ln(\cos x) \), \( \ln(u) \) is outer, and \( u = \cos x \) is inner.
• Differentiate the outer function: \( \frac{d}{du}[\ln u] = \frac{1}{u} \).
• Differentiate the inner function: \( \frac{d}{dx}[\cos x] = -\sin x \).
• Multiply the derivatives together: \( \frac{-\sin x}{\cos x} = -\tan x \).

The chain rule was crucial in determining \( \frac{d}{dx}[\ln(\cos x)] = -\tan x \), which then helped complete the differentiation process of the function.

Product Rule

The product rule is a rule for differentiating expressions where two functions are multiplied together. For any product \( u(x) \cdot v(x) \), the rule states that the derivative is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

This concept was applied in our solution during the differentiation of \( x \ln(\cos x) \), where we treated it as a product of two functions, \( x \) and \( \ln(\cos x) \).

• Identify the two functions: \( u = x \) and \( v = \ln(\cos x) \).
• Differentiate each function: \( u' = 1 \), and using the chain rule for \( v' \), \( v' = -\tan x \).
• Apply the product rule: \( 1 \cdot \ln(\cos x) + x \cdot (-\tan x) \).
• Combine these results: \( \ln(\cos x) - x \tan x \).

By utilizing the product rule, we managed to differentiate the expression accurately, which then allowed us to solve for the derivative of the original function efficiently.