To find the Taylor series of \( f(x) = 2x e^{e^{(3x)^2}} \) around \( a = 0 \), we need to compute the derivatives of \( f(x) \) at \( x = 0 \) and use the Taylor series formula:

\[
f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots
\]

Let's proceed by evaluating the function and its derivatives at \( x = 0 \).

### Step 1: Evaluate \( f(0) \)
\[
f(0) = 2 \cdot 0 \cdot e^{e^{(3 \cdot 0)^2}} = 0
\]

Since \( f(0) = 0 \), the constant term in the Taylor series is zero.

### Step 2: Compute the First Derivative, \( f'(x) \)
Using the product rule for derivatives:
\[
f(x) = 2x \cdot e^{e^{9x^2}}
\]
\[
f'(x) = 2 \cdot e^{e^{9x^2}} + 2x \cdot \frac{d}{dx}\left(e^{e^{9x^2}}\right)
\]

To differentiate \( e^{e^{9x^2}} \) with respect to \( x \), we use the chain rule:
\[
\frac{d}{dx}\left(e^{e^{9x^2}}\right) = e^{e^{9x^2}} \cdot \frac{d}{dx}(e^{9x^2}) = e^{e^{9x^2}} \cdot e^{9x^2} \cdot (18x)
\]

Therefore,
\[
f'(x) = 2 e^{e^{9x^2}} + 2x \cdot e^{e^{9x^2}} \cdot e^{9x^2} \cdot 18x = 2 e^{e^{9x^2}} + 36x^2 e^{e^{9x^2}} e^{9x^2}
\]

Now, evaluate \( f'(0) \):
\[
f'(0) = 2 e^{e^0} = 2e
\]

So, the coefficient of \( x \) in the Taylor series is \( 2e \).

### Step 3: Compute the Second Derivative, \( f''(x) \)
This calculation involves differentiating \( f'(x) \) and becomes quite complex. However, for practical purposes, you might only need the first few terms.

After computing higher derivatives, you

Learn how to compute the Taylor series of the function f(x)=2xe^(e^(3x)^2) around a=0. This detailed solution involves evaluating derivatives, using the product and chain rules, and applying the Taylor series formula. Suitable for undergraduate calculus students.

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