Answer :
Answer:
To find \((f \circ f)'(0)\) where \(f(x) = \tan(x)\), we first need to find the composition of the function \(f \circ f\), which is \(f(f(x)) = \tan(\tan(x))\).
Next, we need to find the derivative of this composed function and then evaluate it at \(x = 0\).
### Step 1: Compute the Derivative
Let \(h(x) = \tan(\tan(x))\).
Using the chain rule, we have:
\[ h'(x) = \frac{d}{dx} \left( \tan(\tan(x)) \right) = \sec^2(\tan(x)) \cdot \frac{d}{dx} (\tan(x)) = \sec^2(\tan(x)) \cdot \sec^2(x) \]
### Step 2: Evaluate the Derivative at \(x = 0\)
Now, we need to evaluate \(h'(x)\) at \(x = 0\).
\[ h'(0) = \sec^2(\tan(0)) \cdot \sec^2(0) \]
Since \(\tan(0) = 0\), we get:
\[ \sec^2(\tan(0)) = \sec^2(0) \]
And since \(\sec(0) = 1\), we get:
\[ \sec^2(0) = 1 \]
Thus:
\[ h'(0) = \sec^2(0) \cdot \sec^2(0) = 1 \cdot 1 = 1 \]
Therefore, \((f \circ f)'(0) = 1\).