Differentiate the following w.r.t. x :
`x^(2)l_(n)(sqrt((x^(2)+9)/(x^(2)+4)))`

Answer

To differentiate the function \( y = x^2 \ln\left(\sqrt{\frac{x^2 + 9}{x^2 + 4}}\right) \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the function
We can simplify the logarithmic expression:
\[ y = x^2 \ln\left(\left(\frac{x^2 + 9}{x^2 + 4}\right)^{1/2}\right) = \frac{1}{2} x^2 \ln\left(\frac{x^2 + 9}{x^2 + 4}\right) \]

### Step 2: Differentiate using the product rule
Now, we will differentiate \( y \) using the product rule. Let \( u = \frac{1}{2} x^2 \) and \( v = \ln\left(\frac{x^2 + 9}{x^2 + 4}\right) \). The product rule states that:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step 3: Calculate \( \frac{du}{dx} \)
Differentiating \( u \):
\[ \frac{du}{dx} = \frac{d}{dx}\left(\frac{1}{2} x^2\right) = x \]

### Step 4: Calculate \( \frac{dv}{dx} \)
Now we differentiate \( v \):
\[ v = \ln\left(\frac{x^2 + 9}{x^2 + 4}\right) \]
Using the chain rule and quotient rule:
\[ \frac{dv}{dx} = \frac{1}{\frac{x^2 + 9}{x^2 + 4}} \cdot \frac{d}{dx}\left(\frac{x^2 + 9}{x^2 + 4}\right) \]
Using the quotient rule:
\[ \frac{d}{dx}\left(\frac{x^2 + 9}{x^2 + 4}\right) = \frac{(x^2 + 4)(2x) - (x^2 + 9)(2x)}{(x^2 + 4)^2} = \frac{2x[(x^2 + 4) - (x^2 + 9)]}{(x^2 + 4)^2} = \frac{2x(-5)}{(x^2 + 4)^2} = \frac{-10x}{(x^2 + 4)^2} \]
Thus,
\[ \frac{dv}{dx} = \frac{(x^2 + 4)}{(x^2 + 9)} \cdot \frac{-10x}{(x^2 + 4)^2} = \frac{-10x}{(x^2 + 9)(x^2 + 4)} \]

### Step 5: Substitute back into the product rule
Now substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the product rule:
\[ \frac{dy}{dx} = \frac{1}{2} x^2 \cdot \frac{-10x}{(x^2 + 9)(x^2 + 4)} + \ln\left(\frac{x^2 + 9}{x^2 + 4}\right) \cdot x \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-5x^3}{(x^2 + 9)(x^2 + 4)} + x \ln\left(\frac{x^2 + 9}{x^2 + 4}\right) \]

### Final Result
Thus, the derivative of the function is:
\[ \frac{dy}{dx} = \frac{-5x^3}{(x^2 + 9)(x^2 + 4)} + x \ln\left(\frac{x^2 + 9}{x^2 + 4}\right) \]

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