**Problem Statement:**Given the function \( f(x) = \int_{-3}^{\sin x} e^{t^2} \, dt \), determine \( f'(x) \).**Explanation:**In this problem, we are given a function \( f(x) \) which is defined as an integral with variable upper limit \(\sin x\). To find the derivative \( f'(x) \), we can use the Fundamental Theorem of Calculus, particularly the Leibniz rule for differentiation under the integral sign when the limits depend on \( x \).**Fundamental Theorem of Calculus:**If \( F(u) = \int_{a}^{u} g(t) \, dt \), then \( \frac{d}{du} F(u) = g(u) \).**Applying the Chain Rule:**Since the upper limit \( \sin x \) depends on \( x \), the chain rule needs to be applied. Let \( u = \sin x \), thus: \[ f(x) = F(\sin x) \quad \text{where} \quad F(u) = \int_{-3}^{u} e^{t^2} \, dt. \]Taking the derivative, we get:\[ f'(x) = \frac{d}{dx}\left( F(\sin x) \right). \]Using the chain rule:\[ f'(x) = F'(\sin x) \cdot \frac{d}{dx} (\sin x). \]From the Fundamental Theorem of Calculus:\[ F'(u) = e^{u^2}. \]Therefore:\[ F'(\sin x) = e^{(\sin x)^2}. \]And since:\[ \frac{d}{dx} (\sin x) = \cos x, \]we have:\[ f'(x) = e^{(\sin x)^2} \cdot \cos x. \]So, by applying these rules, we find:\[ f'(x) = e^{(\sin x)^2} \cos x. \]

Name: **Problem Statement:**Given the function \( f(x) = \int_{-3}^{\sin x}
Uploaded: 2024-03-20T06:47:22.000Z
Channel: Bartleby
Description: **Problem Statement:**Given the function \( f(x) = \int_{-3}^{\sin x} e^{t^2} \, dt \), determine \( f'(x) \).**Explanation:**In this problem, we are given a function \( f(x) \) which is defined as an integral with variable upper limit \(\sin x\). T