$$ f ^ { \prime } ( x ) = e ^ { 2 x } [ 2 \ln x + \frac { 1 } { x } ] $$

Question

UpStudy AI · Accepted Answer

We need to find a function $$ f(x) $$ such that

$$
f'(x) = e^{2x}\left[ 2\ln x + \frac{1}{x} \right].
$$

A common strategy is to look for a function whose derivative, when computed using the product rule, gives us the expression above.

**Step 1. Propose a Candidate Function**

Notice that when differentiating a product of functions, the product rule states:

$$
\frac{d}{dx}\left[u(x)v(x)\right] = u'(x)v(x) + u(x)v'(x).
$$

Let us consider the function

$$
f(x) = e^{2x}\ln x.
$$

**Step 2. Differentiate the Proposed Function**

We differentiate $$ f(x) = e^{2x}\ln x $$ using the product rule. Let

$$
u(x) = e^{2x} \quad \text{and} \quad v(x) = \ln x.
$$

First, compute the derivatives of $$ u(x) $$ and $$ v(x) $$:

$$
u'(x) = \frac{d}{dx}\left(e^{2x}\right) = 2e^{2x},
$$
$$
v'(x) = \frac{d}{dx}\left(\ln x\right) = \frac{1}{x}.
$$

Now apply the product rule:

$$
f'(x) = u'(x)v(x) + u(x)v'(x) = 2e^{2x}\ln x + e^{2x}\cdot\frac{1}{x}.
$$

Factor $$ e^{2x} $$ out:

$$
f'(x) = e^{2x}\left[2\ln x + \frac{1}{x}\right].
$$

This exactly matches the given derivative.

**Step 3. Write the Final Answer**

So the function $$ f(x) $$ is

$$
f(x) = e^{2x}\ln x + C,
$$

where $$ C $$ is an arbitrary constant.

Thus, the antiderivative is

$$
\boxed{f(x) = e^{2x}\ln x + C.}
$$
The function is $$ f(x) = e^{2x}\ln x + C $$.

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