Find `(dy)/(dx)`, when:
`y=(x^(3)sinx)/(e^(x))`

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \frac{x^3 \sin x}{e^x}\), we will use logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Take the natural logarithm of both sides
We start by taking the logarithm of both sides:
\[ \ln y = \ln\left(\frac{x^3 \sin x}{e^x}\right) \]

### Step 2: Apply logarithmic properties
Using the properties of logarithms, we can simplify the right-hand side:
\[ \ln y = \ln(x^3) + \ln(\sin x) - \ln(e^x) \]
This simplifies to:
\[ \ln y = 3 \ln x + \ln(\sin x) - x \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \(x\):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(3 \ln x) + \frac{d}{dx}(\ln(\sin x)) - \frac{d}{dx}(x) \]

### Step 4: Differentiate each term
Calculating the derivatives:
1. \(\frac{d}{dx}(3 \ln x) = \frac{3}{x}\)
2. \(\frac{d}{dx}(\ln(\sin x)) = \frac{\cos x}{\sin x} = \cot x\)
3. \(\frac{d}{dx}(x) = 1\)

Putting it all together, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{3}{x} + \cot x - 1 \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now, we multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = y \left(\frac{3}{x} + \cot x - 1\right) \]

### Step 6: Substitute back for \(y\)
Recall that \(y = \frac{x^3 \sin x}{e^x}\). Substitute this back into the equation:
\[ \frac{dy}{dx} = \frac{x^3 \sin x}{e^x} \left(\frac{3}{x} + \cot x - 1\right) \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{x^3 \sin x}{e^x} \left(\frac{3}{x} + \cot x - 1\right) \]

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