In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = ∫(from 0 to lnx) sin(e^t) dt

검증된 단계별 안내

Identify the function given: \( y = \int_0^{\ln x} \sin(e^t) \, dt \). This is a definite integral with a variable upper limit \( \ln x \).

Recall the Fundamental Theorem of Calculus Part 1, which states that if \( y = \int_a^{g(x)} f(t) \, dt \), then \( \frac{dy}{dx} = f(g(x)) \cdot g'(x) \).

In this problem, \( f(t) = \sin(e^t) \) and the upper limit is \( g(x) = \ln x \). The lower limit is a constant (0), so it does not affect the derivative.

Compute the derivative of the upper limit: \( \frac{d}{dx} (\ln x) = \frac{1}{x} \).

Apply the chain rule: \( \frac{dy}{dx} = \sin(e^{\ln x}) \cdot \frac{1}{x} \). Note that \( e^{\ln x} = x \), so the derivative simplifies to \( \frac{dy}{dx} = \sin(x) \cdot \frac{1}{x} \).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that if a function is defined as an integral with a variable upper limit, its derivative is the integrand evaluated at that limit. It allows us to differentiate an integral with variable limits directly.

Chain Rule

The chain rule is used to differentiate composite functions. When the upper limit of the integral is a function of x (like ln(x)), we must multiply the derivative of the integral's upper limit by the derivative of that function to find the overall derivative.

Properties of Logarithmic and Exponential Functions

Understanding the derivatives of ln(x) and e^t is essential here. The derivative of ln(x) is 1/x, and e^t is its own derivative. These properties help simplify the expression when applying the chain rule and evaluating the integrand.

In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.y = ∫(from 0 to lnx) sin(e^t) dt