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 \( h'(x) \)
- Calculate \( g'(x) = \frac{d}{dx}(e^x) = e^x \).
- 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:
\[ h'(x) = x e^x \]