Question

Find the derivative. \(\frac{d}{d x} \int_{1}^{\ln (x)}\left(4 t+e^{t}\right) d t\)

Short answer

The derivative is \(\frac{4\ln(x)}{x} + 1\).

Step-by-step solution

Recognize the Integral Rule

The problem requires finding the derivative of an integral with a variable upper limit. This situation can be solved using the Fundamental Theorem of Calculus, Part 2, which states that if \(F(x) = \int_{a}^{g(x)} f(t) \, dt\), then \(\frac{d}{dx} F(x) = f(g(x)) \cdot g'(x)\). Here, \(f(t) = 4t + e^t\) and \(g(x) = \ln(x)\).

Differentiate the Upper Limit

Identify and differentiate the upper limit function \(g(x) = \ln(x)\) with respect to \(x\). The derivative \(g'(x)\) is \(\frac{1}{x}\).

Apply the Fundamental Theorem of Calculus

Substitute \(g(x) = \ln(x)\) into the function \(f(t)\). Thus, \(f(g(x)) = f(\ln(x)) = 4\ln(x) + e^{\ln(x)}\). Since \(e^{\ln(x)} = x\), it simplifies to \(4\ln(x) + x\).

Multiply by the Derivative of the Upper Limit

Following the rule, multiply the substituted function \(4\ln(x) + x\) by the derivative of the upper limit \(g'(x)\), which is \(\frac{1}{x}\). Therefore, the expression becomes \(\left(4\ln(x) + x\right) \cdot \frac{1}{x}\).

Simplify the Expression

Distribute the multiplication to get \(\frac{4\ln(x)}{x} + \frac{x}{x}\), which simplifies to \(\frac{4\ln(x)}{x} + 1\). Thus, the derivative is \(\frac{4\ln(x)}{x} + 1\).

Key concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a pivotal concept that bridges the gap between differentiation and integration. It essentially tells us that integration and differentiation are inverse processes. Part 2 of this theorem is particularly useful when you want to find the derivative of an integral with a variable upper limit. Consider the function defined by an integral: \[ F(x) = \int_{a}^{g(x)} f(t) \, dt \] The theorem states that the derivative of this function with respect to \(x\) is \[ \frac{d}{dx} F(x) = f(g(x)) \cdot g'(x) \] This is incredibly helpful because it allows us to differentiate complex-looking integrals by focusing on the function inside the integral, \(f(t)\), and the variable upper limit, \(g(x)\). In the original exercise, the integral is defined with a variable upper limit as \( \ln(x) \), and the integrand is \( 4t + e^t \). Part 2 of the Fundamental Theorem of Calculus gives a straightforward way to tackle such problems by using the values of \(f(g(x))\) and the derivative of \(g(x)\) to compute the original derivative.

Differentiation

Differentiation is a core concept within calculus that involves computing the rate at which a particular function is changing at any point. It concerns itself with slopes of curves, which is often visualized as the tangent line to the function graph at a point. When finding the derivative of a function, particularly one linked with a variable upper limit in an integral, we use differentiation in a specific context. In the problem tackled here, we needed to differentiate the function \(g(x) = \ln(x)\). The derivative of \(\ln(x)\) with respect to \(x\) is \[ g'(x) = \frac{1}{x} \] This calculation is important because it directly multiplies with \(f(g(x))\) from the Fundamental Theorem of Calculus, thereby contributing to our final answer. This connection between differentiation and the theorem underlines how fundamental differentiation tools are applied to obtain the derivative of an integral function.

Integral Calculus

Integral Calculus deals with accumulation and areas, primarily focused on computing the total size, length, area, or even volume under curves. Unlike differentiation, which looks at instant changes, integration accumulates values over an interval.In our specific exercise, we had an integral with an upper limit \( \ln(x) \). The integrand function \(f(t) = 4t + e^t\) determines what is accumulated. However, this integral’s evaluation changes with the upper limit, which leads us to involve the Fundamental Theorem of Calculus for differentiation purposes.Though here we focused on finding the derivative of the integral, integral calculus also lays the foundation for numerous applications:

• Calculating areas and volumes
• Determining total change from a rate of change
• Solving differential equations

By integrating a function over an interval, and then differentiating using the theorem, integral calculus plays a crucial role in finding solutions and making sense of complex mathematical descriptions.