Problem 52 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$y=\int_{\tan x}^{0} \frac{d t}{1+t^{2}}$$

Step-by-Step Solution

Answer

The derivative \( \frac{d y}{d x} \) is \(-1\).

Step 1: Recognize the Integral Form

Given the function \[ y = \int_{\tan x}^{0} \frac{d t}{1+t^{2}} \]we see it involves the integral of a function with respect to \( t \) with variable limits of integration. In this case, the upper limit is a constant, and the lower limit is \( \tan x \).

Step 2: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, when dealing with a definite integral with variable limits, the derivative of the integral with respect to its variable upper limit is the integrand evaluated at that limit, multiplied by the derivative of the limit with respect to \( x \). Here, the integral has the lower limit as a variable, \( \tan x \).

Step 3: Adjust the Variable Limits with Chain Rule

Since the integral has a variable lower limit, we apply the chain rule differently. Reversing the limits changes the integral's sign:\[ y = - \int_{0}^{\tan x} \frac{d t}{1+t^{2}} \]Now, apply the chain rule using the upper limit \( \tan x \). Differentiate:\[ \frac{d y}{d x} = - \frac{1}{1 + (\tan x)^2} \cdot \frac{d}{dx}(\tan x) \]

Step 4: Simplify the Derivative Expression

Recall that the derivative of \( \tan x \) is \( \sec^2 x \). Substitute this into the derivative expression:\[ \frac{d y}{d x} = - \frac{1}{1 + \tan^2 x} \cdot \sec^2 x \]Note that \( 1 + \tan^2 x = \sec^2 x \), so simplify further:\[ \frac{d y}{d x} = -1 \]

Step 5: Conclude with the Final Answer

This simplification shows that the derivative of \( y \) with respect to \( x \) is simply a constant, \( -1 \). Therefore:\[ \frac{d y}{d x} = -1 \]

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the world of derivatives and integrals, two core concepts of calculus. It states that differentiation and integration are inverse operations. This theorem is essential when dealing with integrals that have variable limits of integration.
In our exercise, we see an integral with a variable lower limit. The Fundamental Theorem helps us understand how taking the derivative of such an integral relates to the original function inside the integral.

• Imagine an integral of a function from a fixed point to a variable endpoint; the derivative of this integral, with respect to the variable, is the original function evaluated at that endpoint.
• In the expression, the variable limit is the lower limit, which complicates the application since the typical form is for upper limits. Reversing the limits of integration inverts the sign of the integral, allowing us to apply this fundamental principle.

Recognizing and applying this theorem is crucial in solving such problems effectively.

Chain Rule

The Chain Rule is a fundamental tool in calculus used to differentiate compositions of functions. When you have a function nested inside another, the Chain Rule becomes your best friend.
It suggests that to differentiate a composite function, you need to differentiate the outer function at its inner function and then multiply by the derivative of the inner function.

• In our context, we deal with an integral where the integration limits themselves are functions of another variable.
• The Chain Rule helps differentiate the integral when introducing changes in its limits, particularly when these limits are expressions like \( \tan x \).

Using the Chain Rule correctly helps us transition from our manipulated integral into a derivative, aiding in reaching the final answer. It's like unwrapping layers until you reach the simplest form.

Variable Limits of Integration

Integrals with variable limits introduce an interesting challenge. These integrals are not just numbers but functions dependent on their variable limits.
This means that as the limits change, the value of the integral changes, and these changes are critical when we need to find the derivative of such integrals.

• The key idea is that the integral represents an accumulation function—the accumulation of values from one limit to another.
• When the limits themselves are functions of a variable, like \( \tan x \) in our problem, the integration becomes dynamic, adjusting with every change in \( x \).

To address these changes systematically, we set the problem into a more familiar form—by reversing limits and applying the Fundamental Theorem of Calculus, treating them as compositions of functions, and applying the Chain Rule. This structured approach provides clarity and precision in dealing with variable limits, giving confidence in finding precise solutions.