If `y = (e^x logx)/x^2` then `(dy)/(dx)`

Answer

To find the derivative of the function \( y = \frac{e^x \log x}{x^2} \), we will use the quotient rule and the product rule. Let's go through the steps:

### Step 1: Identify the components of the function
We can rewrite the function as:
\[ y = \frac{u}{v} \]
where \( u = e^x \log x \) and \( v = x^2 \).

### Step 2: Apply the Quotient Rule
The quotient rule states that if \( y = \frac{u}{v} \), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
We need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

### Step 3: Differentiate \( u \) and \( v \)
1. **Differentiate \( v \)**:
\[ v = x^2 \implies \frac{dv}{dx} = 2x \]

2. **Differentiate \( u \)** using the product rule:
\[ u = e^x \log x \]
Let \( a = e^x \) and \( b = \log x \). Then:
\[ \frac{du}{dx} = a \frac{db}{dx} + b \frac{da}{dx} \]
We have:
- \( \frac{da}{dx} = e^x \)
- \( \frac{db}{dx} = \frac{1}{x} \)

Therefore:
\[ \frac{du}{dx} = e^x \cdot \frac{1}{x} + \log x \cdot e^x = e^x \left(\frac{1}{x} + \log x\right) \]

### Step 4: Substitute into the Quotient Rule
Now substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule:
\[ \frac{dy}{dx} = \frac{x^2 \left(e^x \left(\frac{1}{x} + \log x\right)\right) - e^x \log x \cdot 2x}{(x^2)^2} \]

### Step 5: Simplify the expression
This simplifies to:
\[ \frac{dy}{dx} = \frac{x^2 e^x \left(\frac{1}{x} + \log x\right) - 2x e^x \log x}{x^4} \]

Factoring out \( e^x \):
\[ \frac{dy}{dx} = \frac{e^x \left(x^2 \left(\frac{1}{x} + \log x\right) - 2x \log x\right)}{x^4} \]

Now simplifying the expression inside the parentheses:
\[ x^2 \left(\frac{1}{x} + \log x\right) = x + x^2 \log x \]
Thus:
\[ x + x^2 \log x - 2x \log x = x + (x^2 - 2x) \log x \]

### Final Expression
So we have:
\[ \frac{dy}{dx} = \frac{e^x \left(x + (x^2 - 2x) \log x\right)}{x^4} \]