Question

Use part one of the fundamental theorem of calculus to find the derivative of the function.
h(x) =
ex
3 ln(t) dt

1

Answer 1

To find the derivative of the function defined using the fundamental theorem of calculus, we start by interpreting the given function:

\[
h(x) = \int_{1}^{e^x} \ln(t) \, dt
\]

According to the first part of the fundamental theorem of calculus, if \( F(t) \) is an antiderivative of the integrand \( \ln(t) \), then:

\[
\frac{d}{dx} \left( \int_{a}^{g(x)} f(t) \, dt \right) = f(g(x)) \cdot g'(x)
\]

where \( g(x) \) is a differentiable function. In this case, \( g(x) = e^x \) and \( f(t) = \ln(t) \).

### Step 1: Find an Antiderivative
First, we need to find an antiderivative \( F(t) \) for \( f(t) = \ln(t) \).

Using integration by parts:

Let:
- \( u = \ln(t) \) (therefore \( du = \frac{1}{t} dt \))
- \( dv = dt \) (therefore \( v = t \))

Then, by integration by parts:
\[
\int \ln(t) \, dt = t \ln(t) - \int t \cdot \frac{1}{t} \, dt = t \ln(t) - t + C
\]

Thus, we have:
\[
F(t) = t \ln(t) - t
\]

### Step 2: Applying the Fundamental Theorem of Calculus
Now, apply the first part of the fundamental theorem of calculus to find the derivative of \( h(x) \):
\[
h'(x) = \frac{d}{dx} \left( \int_{1}^{e^x} \ln(t) \, dt \right) = \ln(e^x) \cdot \frac{d}{dx}(e^x)
\]

### Step 3: Calculate \( h'(x) \)
1. Calculate \( g'(x) = \frac{d}{dx}(e^x) = e^x \).
2. The function \( \ln(e^x) = x \).

Putting it all together, we have:
\[
h'(x) = \ln(e^x) \cdot e^x = x \cdot e^x
\]

### Final Answer
Thus, the derivative of the function \( h(x) \) is:

\[
h'(x) = x e^x
\]