`(e^(x) sinx)/(secx)`

Answer

To differentiate the expression \( y = \frac{e^x \sin x}{\sec x} \), we can simplify it first and then apply the product rule. Here’s a step-by-step solution:

### Step 1: Simplify the expression
We know that \( \sec x = \frac{1}{\cos x} \). Therefore, we can rewrite the expression as:
\[ y = e^x \sin x \cdot \cos x \]

### Step 2: Identify the functions
Let \( u = e^x \sin x \) and \( v = \cos x \). We will differentiate \( y = u \cdot v \) using the product rule.

### Step 3: Apply the product rule
The product rule states that if \( y = u \cdot v \), then:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
We need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

### Step 4: Differentiate \( u \)
To differentiate \( u = e^x \sin x \), we apply the product rule again:
\[ \frac{du}{dx} = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x) \]

### Step 5: Differentiate \( v \)
Now, differentiate \( v = \cos x \):
\[ \frac{dv}{dx} = -\sin x \]

### Step 6: Substitute back into the product rule
Now substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) back into the product rule:
\[ \frac{dy}{dx} = (e^x \sin x)(-\sin x) + (\cos x)(e^x (\sin x + \cos x)) \]

### Step 7: Simplify the expression
Now simplify the expression:
\[ \frac{dy}{dx} = -e^x \sin^2 x + e^x \cos x (\sin x + \cos x) \]
\[ = -e^x \sin^2 x + e^x \cos x \sin x + e^x \cos^2 x \]

### Step 8: Factor out \( e^x \)
Now we can factor out \( e^x \):
\[ \frac{dy}{dx} = e^x (-\sin^2 x + \cos x \sin x + \cos^2 x) \]

### Step 9: Use the Pythagorean identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \frac{dy}{dx} = e^x (1 + \cos x \sin x - \sin^2 x) \]

### Final Answer
Thus, the derivative of the given expression is:
\[ \frac{dy}{dx} = e^x (1 + \cos x \sin x - \sin^2 x) \]