Find `(dy)/(dx)`, when:
`y=((x+1)^(2)sqrt(x-1))/((x+4)^(3).e^(x))`

Explanation

To find \(\frac{dy}{dx}\) for the function

\[ y = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}, \]

we will use logarithmic differentiation to simplify the process.

### Step 1: Take the natural logarithm of both sides

Start by taking the natural logarithm of \(y\):

\[ \ln y = \ln \left( \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \right). \]

### Step 2: Apply logarithmic properties

Using the properties of logarithms, we can separate the terms:

\[ \ln y = \ln((x+1)^2) + \ln(\sqrt{x-1}) - \ln((x+4)^3) - \ln(e^x). \]

This simplifies to:

\[ \ln y = 2 \ln(x+1) + \frac{1}{2} \ln(x-1) - 3 \ln(x+4) - x. \]

### Step 3: Differentiate both sides with respect to \(x\)

Now, we differentiate both sides with respect to \(x\):

Using implicit differentiation on the left side:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1. \]

### Step 4: Solve for \(\frac{dy}{dx}\)

Now, multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = y \left( \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right). \]

### Step 5: Substitute back for \(y\)

Substituting back the expression for \(y\):

\[ \frac{dy}{dx} = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \left( \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right). \]

### Final Answer

Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \left( \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right). \]

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