If y = (x x ) x , then which one of the following is correct ?
Finding the Derivative of \(y = (x^x)^x\)
The problem asks us to find the derivative of the function \(y = (x^x)^x\) and determine which of the given differential equations it satisfies. This type of function, where both the base and the exponent involve the variable \(x\), requires a special technique called logarithmic differentiation.
Simplifying the Function
Before differentiating, we can simplify the given function using the exponent rule \((a^b)^c = a^{b \cdot c}\).
The function is \(y = (x^x)^x\). Here, \(a = x\), \(b = x\), and \(c = x\). So, we can write:
\[y = x^{x \cdot x} = x^{x^2}\]
Now the function is in the form \(y = f(x)^{g(x)}\), where \(f(x) = x\) and \(g(x) = x^2\). We will use logarithmic differentiation on this simplified form.
Applying Logarithmic Differentiation
Logarithmic differentiation involves taking the natural logarithm of both sides of the equation and then differentiating implicitly with respect to \(x\).
Step 1: Take the natural logarithm of both sides
\[\ln y = \ln(x^{x^2})\]
Step 2: Use logarithm properties to simplify the right side
Using the property \(\ln(a^b) = b \ln a\), we bring the exponent \(x^2\) down:
\[\ln y = x^2 \ln x\]
Step 3: Differentiate both sides with respect to \(x\)
We differentiate \(\ln y\) with respect to \(x\) using the chain rule, and \(x^2 \ln x\) with respect to \(x\) using the product rule.
Recall the rules:
- Chain Rule: \(\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}\)
- Product Rule: \(\frac{d}{dx}(u \cdot v) = u'v + uv'\)
Let \(u = x^2\) and \(v = \ln x\). Then \(u' = \frac{d}{dx}(x^2) = 2x\) and \(v' = \frac{d}{dx}(\ln x) = \frac{1}{x}\).
Applying the differentiation:
\[\frac{d}{dx}(\ln y) = \frac{d}{dx}(x^2 \ln x)\]
\[\frac{1}{y} \frac{dy}{dx} = (2x)(\ln x) + (x^2)\left(\frac{1}{x}\right)\]
\[\frac{1}{y} \frac{dy}{dx} = 2x \ln x + \frac{x^2}{x}\]
\[\frac{1}{y} \frac{dy}{dx} = 2x \ln x + x\]
We can factor out \(x\) from the terms on the right side:
\[\frac{1}{y} \frac{dy}{dx} = x(2 \ln x + 1)\]
Step 4: Solve for \(\frac{dy}{dx}\)
Multiply both sides by \(y\):
\[\frac{dy}{dx} = y \cdot x(2 \ln x + 1)\]
Rearranging the terms slightly:
\[\frac{dy}{dx} = xy(1 + 2 \ln x)\]
Comparing the Result with Options
We have found the derivative \(\frac{dy}{dx} = xy(1 + 2 \ln x)\). Now let's look at the given options, which are in the form of differential equations.
The equation we found can be rewritten by moving the term \(xy(1 + 2 \ln x)\) to the left side:
\[\frac{dy}{dx} - xy(1 + 2 \ln x) = 0\]
Let's compare this with the given options:
| Option | Equation |
|---|---|
| 1 | \(\frac{dy}{dx}\) + xy(1 + 2 ln x) = 0 |
| 2 | \(\frac{dy}{dx}\) − xy(1 + 2 ln x) = 0 |
| 3 | \(\frac{dy}{dx}\) − 2xy(1 + ln x) = 0 |
| 4 | \(\frac{dy}{dx}\) + 2xy(1 + ln x) = 0 |
Comparing our derived equation \(\frac{dy}{dx} - xy(1 + 2 \ln x) = 0\) with the options, we see that it exactly matches Option 2.
Summary of Differentiation Steps
To find the derivative of \(y = (x^x)^x = x^{x^2}\):
- Simplify the exponent: \(y = x^{x^2}\).
- Take the natural log: \(\ln y = \ln(x^{x^2}) = x^2 \ln x\).
- Differentiate implicitly w.r.t. \(x\): \(\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}(x^2 \ln x)\).
- Apply product rule: \(\frac{1}{y}\frac{dy}{dx} = 2x \ln x + x^2 \left(\frac{1}{x}\right) = 2x \ln x + x\).
- Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = y(2x \ln x + x) = yx(2 \ln x + 1)\).
- Substitute \(y = x^{x^2}\): \(\frac{dy}{dx} = x^{x^2} x (1 + 2 \ln x) = x^{x^2+1}(1 + 2 \ln x)\). (This form is also correct, but the options use \(y\)).
- Using the form with \(y\): \(\frac{dy}{dx} = xy(1 + 2 \ln x)\).
- Rearrange to match option format: \(\frac{dy}{dx} - xy(1 + 2 \ln x) = 0\).
Revision Table: Key Concepts for Differentiation
| Concept | Description | Application in this Problem |
|---|---|---|
| Exponent Rule: \((a^b)^c = a^{bc}\) | Simplifies nested exponents. | Used to rewrite \((x^x)^x\) as \(x^{x^2}\). |
| Logarithmic Differentiation | Method used for functions \(y=f(x)^{g(x)}\) or complex products/quotients by taking the log first. | Essential because \(y=x^{x^2}\) has a variable base and exponent. |
| Logarithm Property: \(\ln(a^b) = b \ln a\) | Brings the exponent down as a multiplier. | Used on \(\ln(x^{x^2})\) to get \(x^2 \ln x\). |
| Chain Rule | Used when differentiating a composite function, e.g., \(\ln(y)\) with respect to \(x\). | Applied to \(\frac{d}{dx}(\ln y)\) resulting in \(\frac{1}{y}\frac{dy}{dx}\). |
| Product Rule: \((uv)' = u'v + uv'\) | Used to differentiate the product of two functions. | Applied to differentiate \(x^2 \ln x\). |
| Basic Derivatives | Derivatives of standard functions like \(x^n\) and \(\ln x\). | Used for \(\frac{d}{dx}(x^2) = 2x\) and \(\frac{d}{dx}(\ln x) = \frac{1}{x}\). |
Additional Information: When to Use Logarithmic Differentiation
Logarithmic differentiation is particularly useful in the following cases:
- Functions of the form \(y = f(x)^{g(x)}\), like \(x^x\), \((\sin x)^x\), etc. These cannot be differentiated using simple power rule \((x^n \to nx^{n-1})\) or exponential rule \((a^x \to a^x \ln a)\).
- Functions that are complicated products or quotients, especially when involving many terms. Taking the logarithm first can turn products into sums and quotients into differences, making differentiation simpler. For example, differentiating \(y = \frac{(x+1)^3 \sqrt{x-2}}{(x^2+4)^{1/3}}\).
The process generally involves:
- Taking the natural logarithm of both sides.
- Simplifying using logarithm properties.
- Differentiating implicitly with respect to \(x\).
- Solving for \(\frac{dy}{dx}\).
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