Derivative of Integral: Calculus Solution

To find the derivative of the given function $y = \int_{0}^{\ln x} \sin e^{t} dt$, we need to apply the Fundamental Theorem of Calculus (Part 1) combined with the Chain Rule.

Applying Calculus Theorems

1. Let $u = \ln x$. Then the integral becomes $y = \int_{0}^{u} \sin e^{t} dt$.
2. According to the Fundamental Theorem of Calculus Part 1, if $y = \int_{a}^{u} f(t) dt$, then $\frac{dy}{du} = f(u)$. In this case, $f(t) = \sin e^{t}$, so $\frac{dy}{du} = \sin e^{u}$.
3. We also need the derivative of $u$ with respect to $x$: $\frac{du}{dx} = \frac{d}{dx}(\ln x) = \frac{1}{x}$.
4. Using the Chain Rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.
5. Substitute the results from steps 2 and 3: $\frac{dy}{dx} = (\sin e^{u}) \cdot \left(\frac{1}{x}\right)$.
6. Substitute $u = \ln x$ back into the equation: $\frac{dy}{dx} = \left(\sin e^{\ln x}\right) \cdot \left(\frac{1}{x}\right)$.
7. Simplify the expression using the property $e^{\ln x} = x$: $\frac{dy}{dx} = (\sin x) \cdot \left(\frac{1}{x}\right)$.
8. The final derivative is $\frac{dy}{dx} = \frac{\sin x}{x}$.

Therefore, the derivative of $y = \int_{0}^{\ln x} \sin e^{t} dt$ is $\frac{\sin x}{x}$.