`x=e^t (sin t + cos t ),y=e^t(sin t -cos t)`

Answer

To find \(\frac{dy}{dx}\) for the given parametric equations \(x = e^t (\sin t + \cos t)\) and \(y = e^t (\sin t - \cos t)\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(t\)

We start by differentiating \(x\):
\[ x = e^t (\sin t + \cos t) \]
Using the product rule, we have:
\[ \frac{dx}{dt} = \frac{d}{dt}(e^t) \cdot (\sin t + \cos t) + e^t \cdot \frac{d}{dt}(\sin t + \cos t) \]
Calculating the derivatives:
\[ \frac{d}{dt}(e^t) = e^t \]
\[ \frac{d}{dt}(\sin t + \cos t) = \cos t - \sin t \]
Now substituting back:
\[ \frac{dx}{dt} = e^t (\sin t + \cos t) + e^t (\cos t - \sin t) \]
Combining like terms:
\[ \frac{dx}{dt} = e^t [(\sin t + \cos t) + (\cos t - \sin t)] = e^t [2\cos t] \]

### Step 2: Differentiate \(y\) with respect to \(t\)

Next, we differentiate \(y\):
\[ y = e^t (\sin t - \cos t) \]
Using the product rule again:
\[ \frac{dy}{dt} = \frac{d}{dt}(e^t) \cdot (\sin t - \cos t) + e^t \cdot \frac{d}{dt}(\sin t - \cos t) \]
Calculating the derivatives:
\[ \frac{d}{dt}(\sin t - \cos t) = \cos t + \sin t \]
Now substituting back:
\[ \frac{dy}{dt} = e^t (\sin t - \cos t) + e^t (\cos t + \sin t) \]
Combining like terms:
\[ \frac{dy}{dt} = e^t [(\sin t - \cos t) + (\cos t + \sin t)] = e^t [2\sin t] \]

### Step 3: Find \(\frac{dy}{dx}\)

Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{e^t (2\sin t)}{e^t (2\cos t)} \]
The \(e^t\) terms cancel out:
\[ \frac{dy}{dx} = \frac{2\sin t}{2\cos t} = \frac{\sin t}{\cos t} = \tan t \]

### Final Answer
\[ \frac{dy}{dx} = \tan t \]