Find `dy/dx,"where "x=t^(3)+1/t andy=(t+t^(2))^(3)`.

Answer

To find \(\frac{dy}{dx}\) where \(x = t^3 + \frac{1}{t}\) and \(y = (t + t^2)^3\), we will use the chain rule. The steps are as follows:

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

Given:
\[ y = (t + t^2)^3 \]

Using the chain rule:
\[ \frac{dy}{dt} = 3(t + t^2)^2 \cdot \frac{d}{dt}(t + t^2) \]

Now, differentiate \(t + t^2\):
\[ \frac{d}{dt}(t + t^2) = 1 + 2t \]

Thus,
\[ \frac{dy}{dt} = 3(t + t^2)^2 \cdot (1 + 2t) \]

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

Given:
\[ x = t^3 + \frac{1}{t} \]

Differentiate \(x\):
\[ \frac{dx}{dt} = 3t^2 - \frac{1}{t^2} \]

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{3(t + t^2)^2 \cdot (1 + 2t)}{3t^2 - \frac{1}{t^2}} \]

### Step 4: Simplify the expression

To simplify \(\frac{dy}{dx}\):
1. The numerator is \(3(t + t^2)^2(1 + 2t)\).
2. The denominator can be rewritten as:
\[ 3t^2 - \frac{1}{t^2} = \frac{3t^4 - 1}{t^2} \]

Thus:
\[ \frac{dy}{dx} = \frac{3(t + t^2)^2(1 + 2t) \cdot t^2}{3t^4 - 1} \]

### Final Answer
\[ \frac{dy}{dx} = \frac{3t^2(t + t^2)^2(1 + 2t)}{3t^4 - 1} \]

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