Find `(dy)/(dx)`, when:
`y=e^(x)sin^(3)xcos^(4)x`

Explanation

To find \(\frac{dy}{dx}\) when \(y = e^x \sin^3 x \cos^4 x\), we will use logarithmic differentiation. Here are the steps:

### Step 1: Write the function
We start with the function:
\[ y = e^x \sin^3 x \cos^4 x \]

### Step 2: Take the natural logarithm of both sides
Taking the natural logarithm on both sides gives:
\[ \ln y = \ln(e^x \sin^3 x \cos^4 x) \]

### Step 3: Use properties of logarithms
Using the properties of logarithms, we can expand the right-hand side:
\[ \ln y = \ln(e^x) + \ln(\sin^3 x) + \ln(\cos^4 x) \]
This simplifies to:
\[ \ln y = x + 3 \ln(\sin x) + 4 \ln(\cos x) \]

### Step 4: Differentiate both sides
Now, we differentiate both sides with respect to \(x\):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x) + 3 \frac{d}{dx}(\ln(\sin x)) + 4 \frac{d}{dx}(\ln(\cos x)) \]

### Step 5: Differentiate the right-hand side
Calculating the derivatives:
- The derivative of \(x\) is \(1\).
- The derivative of \(\ln(\sin x)\) is \(\cot x\).
- The derivative of \(\ln(\cos x)\) is \(-\tan x\).

Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = 1 + 3 \cot x - 4 \tan x \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Now, we multiply both sides by \(y\):
\[ \frac{dy}{dx} = y(1 + 3 \cot x - 4 \tan x) \]

### Step 7: Substitute back for \(y\)
Substituting back the expression for \(y\):
\[ \frac{dy}{dx} = e^x \sin^3 x \cos^4 x (1 + 3 \cot x - 4 \tan x) \]

### Final Answer
Thus, the final result is:
\[ \frac{dy}{dx} = e^x \sin^3 x \cos^4 x (1 + 3 \cot x - 4 \tan x) \]

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