To solve this problem using the Fundamental Theorem of Calculus (FTC), we need to compute the derivative $h'(x)$ of the given function:

$h(x) = \int_1^{e^x} \ln t \, dt.$

Step-by-Step Solution:

1. Recall the Fundamental Theorem of Calculus, Part II:

If $F(x) = \int_a^{g(x)} f(t) \, dt$, then the derivative is: $F'(x) = f(g(x)) \cdot g'(x).$

2. Apply the FTC to $h(x)$:

Here, $f(t) = \ln t$, and the upper limit of the integral is $g(x) = e^x$. So: $h'(x) = f(g(x)) \cdot g'(x).$

Substitute $f(t) = \ln t$ and $g(x) = e^x$: $h'(x) = \ln(e^x) \cdot \frac{d}{dx}(e^x).$

3. Simplify the terms:

• $\ln(e^x) = x$ (since $\ln(e^x) = x$ by the logarithm property).
• $\frac{d}{dx}(e^x) = e^x$.

Thus: $h'(x) = x \cdot e^x.$

Final Answer:

$h'(x) = x e^x.$

Would you like further clarification on any step or a deeper explanation?

Related Questions:

1. What is the Fundamental Theorem of Calculus and how is it applied here?
2. How do you compute derivatives of integrals with variable bounds?
3. What are some applications of this type of derivative in real-world contexts?
4. How can you extend this method to solve integrals with more complex bounds?
5. What happens if $\ln t$ were replaced by a different function, such as $t^2$?

Tip:

When taking the derivative of an integral with a variable upper bound, always identify $g(x)$, differentiate it, and evaluate the integrand at $g(x)$.