If y = cos x ⋅ cos 4x ⋅ cos 8x, then what is \(\rm \frac{1}{y}\frac{dy}{dx}\) at \(\rm x = \frac{\pi}{4}\) equal to?
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Understanding the Problem: Finding the Logarithmic Derivative
The question asks us to find the value of the expression \(\rm \frac{1}{y}\frac{dy}{dx}\) for the given function \(y = \cos x \cdot \cos 4x \cdot \cos 8x\) at a specific point, \(\rm x = \frac{\pi}{4}\). The term \(\rm \frac{1}{y}\frac{dy}{dx}\) is known as the logarithmic derivative of the function \(y\).
To find the logarithmic derivative of a function that is a product of several terms, it is usually easiest to use the technique of logarithmic differentiation.
Steps for Logarithmic Differentiation
Logarithmic differentiation involves the following steps:
- Take the natural logarithm (\(\ln\)) of both sides of the equation \(y = f(x)\).
- Use the properties of logarithms to simplify the right-hand side, especially \(\ln(ab) = \ln a + \ln b\).
- Differentiate both sides of the equation with respect to \(x\). The left side will become \(\frac{1}{y}\frac{dy}{dx}\) by the chain rule.
- Solve for \(\frac{dy}{dx}\) if needed, but in this problem, we only need \(\frac{1}{y}\frac{dy}{dx}\).
- Substitute the given value of \(x\) into the resulting expression to find the numerical value.
Applying Logarithmic Differentiation to \(y = \cos x \cdot \cos 4x \cdot \cos 8x\)
Let's apply these steps to the given function \(y = \cos x \cdot \cos 4x \cdot \cos 8x\).
Step 1: Take the Natural Logarithm
Take the natural logarithm of both sides:
\(\ln y = \ln (\cos x \cdot \cos 4x \cdot \cos 8x)\)
Step 2: Simplify using Logarithm Properties
Using the property \(\ln(abc) = \ln a + \ln b + \ln c\), we simplify the right side:
\(\ln y = \ln (\cos x) + \ln (\cos 4x) + \ln (\cos 8x)\)
Step 3: Differentiate with respect to \(x\)
Now, differentiate both sides with respect to \(x\). Remember that \(\frac{d}{dx}(\ln u) = \frac{1}{u}\frac{du}{dx}\).
\(\frac{d}{dx}(\ln y) = \frac{d}{dx}(\ln (\cos x)) + \frac{d}{dx}(\ln (\cos 4x)) + \frac{d}{dx}(\ln (\cos 8x))\)
Applying the chain rule for each term on the right side:
- For \(\ln(\cos x)\): \(\frac{d}{dx}(\ln (\cos x)) = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x) = \frac{1}{\cos x} (-\sin x) = -\frac{\sin x}{\cos x} = -\tan x\)
- For \(\ln(\cos 4x)\): \(\frac{d}{dx}(\ln (\cos 4x)) = \frac{1}{\cos 4x} \cdot \frac{d}{dx}(\cos 4x) = \frac{1}{\cos 4x} (-\sin 4x \cdot 4) = -4\frac{\sin 4x}{\cos 4x} = -4\tan 4x\)
- For \(\ln(\cos 8x)\): \(\frac{d}{dx}(\ln (\cos 8x)) = \frac{1}{\cos 8x} \cdot \frac{d}{dx}(\cos 8x) = \frac{1}{\cos 8x} (-\sin 8x \cdot 8) = -8\frac{\sin 8x}{\cos 8x} = -8\tan 8x\)
So, the derivative of the left side is \(\frac{1}{y}\frac{dy}{dx}\), and the derivative of the right side is the sum of the derivatives of the individual terms:
\(\frac{1}{y}\frac{dy}{dx} = -\tan x - 4\tan 4x - 8\tan 8x\)
Evaluating the Logarithmic Derivative at \(\rm x = \frac{\pi}{4}\)
Now we need to find the value of the expression \(\frac{1}{y}\frac{dy}{dx}\) at \(\rm x = \frac{\pi}{4}\). Substitute \(\rm x = \frac{\pi}{4}\) into the derived expression:
\(\left.\frac{1}{y}\frac{dy}{dx}\right|_{x=\frac{\pi}{4}} = -\tan\left(\frac{\pi}{4}\right) - 4\tan\left(4 \cdot \frac{\pi}{4}\right) - 8\tan\left(8 \cdot \frac{\pi}{4}\right)\)
Simplify the angles:
\(\left.\frac{1}{y}\frac{dy}{dx}\right|_{x=\frac{\pi}{4}} = -\tan\left(\frac{\pi}{4}\right) - 4\tan\left(\pi\right) - 8\tan\left(2\pi\right)\)
Evaluating Trigonometric Functions
We need the values of the tangent function at these specific angles:
- \(\tan\left(\frac{\pi}{4}\right) = 1\)
- \(\tan\left(\pi\right) = 0\)
- \(\tan\left(2\pi\right) = 0\) (Since the tangent function has a period of \(\pi\), \(\tan(2\pi) = \tan(0) = 0\))
Final Calculation
Substitute these values back into the expression:
\(\left.\frac{1}{y}\frac{dy}{dx}\right|_{x=\frac{\pi}{4}} = -(1) - 4(0) - 8(0)\)
\(\left.\frac{1}{y}\frac{dy}{dx}\right|_{x=\frac{\pi}{4}} = -1 - 0 - 0\)
\(\left.\frac{1}{y}\frac{dy}{dx}\right|_{x=\frac{\pi}{4}} = -1\)
Thus, the value of \(\rm \frac{1}{y}\frac{dy}{dx}\) at \(\rm x = \frac{\pi}{4}\) is -1.
Summary of Logarithmic Derivative Calculation
| Step | Action | Result |
|---|---|---|
| 1 | Given function y | \(y = \cos x \cdot \cos 4x \cdot \cos 8x\) |
| 2 | Take ln on both sides | \(\ln y = \ln(\cos x) + \ln(\cos 4x) + \ln(\cos 8x)\) |
| 3 | Differentiate with respect to x | \(\frac{1}{y}\frac{dy}{dx} = -\tan x - 4\tan 4x - 8\tan 8x\) |
| 4 | Substitute \(x = \frac{\pi}{4}\) | \(\left.\frac{1}{y}\frac{dy}{dx}\right|_{x=\frac{\pi}{4}} = -\tan(\frac{\pi}{4}) - 4\tan(\pi) - 8\tan(2\pi)\) |
| 5 | Evaluate trigonometric terms | \(\tan(\frac{\pi}{4}) = 1, \tan(\pi) = 0, \tan(2\pi) = 0\) |
| 6 | Calculate final value | \(-1 - 4(0) - 8(0) = -1\) |
Revision Table: Key Concepts
| Concept | Description | Formula/Example |
|---|---|---|
| Logarithmic Derivative | The derivative of the logarithm of a function; equal to \(\frac{f'(x)}{f(x)}\) or \(\frac{1}{y}\frac{dy}{dx}\). Useful for differentiating products, quotients, and powers. | \(\frac{d}{dx}(\ln f(x)) = \frac{f'(x)}{f(x)}\) |
| Properties of Logarithms | Rules for simplifying logarithmic expressions. | \(\ln(ab) = \ln a + \ln b\) \(\ln(\frac{a}{b}) = \ln a - \ln b\) \(\ln(a^n) = n \ln a\) |
| Chain Rule | A rule for differentiating composite functions. If \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). | \(\frac{d}{dx}(\ln(\cos x)) = \frac{1}{\cos x} \cdot \frac{d}{dx}(\cos x)\) |
| Trigonometric Values | Standard values of trigonometric functions at common angles like \(\frac{\pi}{4}\), \(\pi\), \(2\pi\). | \(\tan(\frac{\pi}{4}) = 1\) \(\tan(\pi) = 0\) |
Additional Information: Why Logarithmic Differentiation?
Logarithmic differentiation is particularly useful when dealing with functions that are complicated products, quotients, or when variables appear in exponents (like \(y = x^x\)).
Consider differentiating \(y = \cos x \cdot \cos 4x \cdot \cos 8x\) directly using the product rule. The product rule for three functions \(u \cdot v \cdot w\) is \((uvw)' = u'vw + uv'w + uvw'\). This would involve calculating the derivative of each cosine term and then combining them, which is much more cumbersome than the logarithmic approach.
By taking the logarithm, the product is converted into a sum, and differentiation of a sum is simpler than differentiation of a product. This transformation simplifies the differentiation process significantly for complex functions.
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