If y =log `(1+ 2t^(2)+t^(4)), x= tan^(-1) ` t find `(d^(2)y)/(dx^(2))`

Answer

To find the second derivative \(\frac{d^2y}{dx^2}\) given \(y = \log(1 + 2t^2 + t^4)\) and \(x = \tan^{-1}(t)\), we will follow these steps:

### Step 1: Simplify \(y\)
We start with the expression for \(y\):
\[ y = \log(1 + 2t^2 + t^4) \]
Notice that we can factor the expression inside the logarithm:
\[ 1 + 2t^2 + t^4 = (t^2 + 1)^2 \]
Thus, we can rewrite \(y\) as:
\[ y = \log((t^2 + 1)^2) = 2 \log(t^2 + 1) \]

### Step 2: Differentiate \(y\) with respect to \(t\)
Now we find the first derivative \(\frac{dy}{dt}\):
\[ \frac{dy}{dt} = 2 \cdot \frac{1}{t^2 + 1} \cdot \frac{d}{dt}(t^2 + 1) = 2 \cdot \frac{1}{t^2 + 1} \cdot 2t = \frac{4t}{t^2 + 1} \]

### Step 3: Differentiate \(x\) with respect to \(t\)
Next, we differentiate \(x\):
\[ x = \tan^{-1}(t) \]
The derivative is:
\[ \frac{dx}{dt} = \frac{1}{1 + t^2} \]

### Step 4: Find \(\frac{dy}{dx}\)
Using the chain rule, we can find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{4t}{t^2 + 1}}{\frac{1}{1 + t^2}} = 4t \]

### Step 5: Differentiate \(\frac{dy}{dx}\) with respect to \(t\)
Now we need to find the second derivative \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} \]
First, we differentiate \(\frac{dy}{dx} = 4t\):
\[ \frac{d}{dt}(4t) = 4 \]

### Step 6: Find \(\frac{dt}{dx}\)
We already have \(\frac{dx}{dt} = \frac{1}{1 + t^2}\), so:
\[ \frac{dt}{dx} = 1 + t^2 \]

### Step 7: Combine to find \(\frac{d^2y}{dx^2}\)
Now, substituting back:
\[ \frac{d^2y}{dx^2} = 4 \cdot (1 + t^2) \]

Thus, the final answer is:
\[ \frac{d^2y}{dx^2} = 4(1 + t^2) \]

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