Find `dy/dx`, when :
`x=e^(t)(sint+cost)andy=e^(t)(sint-cost)`.

Answer

To find \(\frac{dy}{dx}\) given the 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 \(y\) with respect to \(t\)

We start with the equation for \(y\):
\[ y = e^t (\sin t - \cos t) \]

Using the product rule, we differentiate \(y\):
\[ \frac{dy}{dt} = \frac{d}{dt}(e^t) \cdot (\sin t - \cos t) + e^t \cdot \frac{d}{dt}(\sin t - \cos t) \]

Calculating each part:
1. \(\frac{d}{dt}(e^t) = e^t\)
2. \(\frac{d}{dt}(\sin t - \cos t) = \cos t + \sin t\)

So we have:
\[ \frac{dy}{dt} = e^t (\sin t - \cos t) + e^t (\cos t + \sin t) \]
\[ \frac{dy}{dt} = e^t ((\sin t - \cos t) + (\cos t + \sin t)) = e^t (2\sin t) \]

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

Now we differentiate \(x\):
\[ x = e^t (\sin t + \cos t) \]

Using the product rule again:
\[ \frac{dx}{dt} = \frac{d}{dt}(e^t) \cdot (\sin t + \cos t) + e^t \cdot \frac{d}{dt}(\sin t + \cos t) \]

Calculating each part:
1. \(\frac{d}{dt}(e^t) = e^t\)
2. \(\frac{d}{dt}(\sin t + \cos t) = \cos t - \sin t\)

So we have:
\[ \frac{dx}{dt} = e^t (\sin t + \cos t) + e^t (\cos t - \sin t) \]
\[ \frac{dx}{dt} = e^t ((\sin t + \cos t) + (\cos t - \sin t)) = e^t (2\cos t) \]

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

Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{e^t (2\sin t)}{e^t (2\cos t)} \]

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:
\[ \frac{dy}{dx} = \tan t \]

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