Problem 31 - Calculus: Early Transcendental Functions

Question

find the derivative \(f^{\prime}(x)\) \(f(x)=\int_{x}^{-1} \ln \left(t^{2}+1\right) d t\)

Step-by-Step Solution

Answer

The derivative of the given function is \(f'(x) = -\ln(x^2 + 1) + \ln(2)\)

Step 1: Identify the integrand

The integrand in the given function is \(\ln(t^2 + 1)\). Thus, \(f(t) = \ln(t^2 + 1)\).

Step 2: Apply the Fundamental Theorem of Calculus Part 1

According to the Fundamental Theorem of Calculus, the derivative of \(f(x)\) is given by \(f'(x) = -f(x) + f(-1)\). The negative sign is because the upper and lower limits of the integral are reversed.

Step 3: Substitute the limits

Substitute \(x\) and \(-1\) into the function \(f(t)\) to get \(f'(x) = -f(x) + f(-1) = -\ln(x^2 + 1) + \ln(((-1)^2 + 1)) = -\ln(x^2 + 1) + \ln(2)\)

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a principle that bridges the concepts of differentiation and integration, two core operations in calculus. It comes in two parts. The first part provides a way to evaluate a definite integral. It states that if a function 'f' is continuous on the interval [a, b] and 'F' is a function whose derivative is 'f', then the integral of 'f' from a to b is equal to 'F(b) - F(a)'.

The second part, often applied in finding derivatives of integral functions, tells us that if 'f' is a continuous function on an interval, then the function 'F' given by the integral from a to 'x' of 'f(t)dt' is also continuous on that interval. Moreover, 'F' is differentiable, and its derivative is 'f(x)'. This means we can find the derivative of an integral function directly by examining its integrand and the limits of integration. In the case of our exercise, this theorem allows us to establish a direct relationship between the derivative of the integral expression and the original integrand.

Integrand

The integrand is the function we want to integrate and is found within the integral sign. It is the 'heart' of the integral, expressing the function that is being summed over a specific interval. In our given problem, the integrand is \( \ln(t^2 + 1) \), where \(t\) is the variable of integration. Understanding the nature of the integrand is crucial when applying the Fundamental Theorem of Calculus, as the derivative of the integral is largely dependent on the form and behavior of this integrand over the interval in question.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. Importantly, the natural logarithm of a number is the power to which \(e\) would have to be raised to equal that number. For instance, \( \ln(e) = 1 \) because \(e^1 = e\). In the context of integration and differentiation, the properties of the natural logarithm are particularly useful; for example, its derivative \(\frac{d}{dx} \ln(x) = \frac{1}{x}\), which is a direct consequence of its definition. This is relevant as our integral involves the natural logarithm of \(t^2 + 1\), affecting how we find the derivative of the integral through the Fundamental Theorem of Calculus.

Limits of Integration

Limits of integration define the interval over which integration is performed. They are the 'a' and 'b' in the integral \( \int_a^b f(t) dt \), where 'a' is the lower limit, and 'b' is the upper limit. These limits can be numbers, variables, or infinity. In the problem presented, the limits of integration are 'x' and '-1'. These values are crucial for determining the outcome of the definite integral.

When applying the Fundamental Theorem of Calculus to find the derivative of an integral with variable limits, as in our problem, the derivative with respect to 'x' would take into account these changing boundaries. Therefore, special attention must be paid when the limits involve variables, as the derivative will need to reflect the rate of change not just in the integrand but also in these boundary values.