Problem 104 - Calculus Early Transcendentals

Question

$$\frac{d}{d t}\left(\int_{0}^{t} \frac{d x}{1+x^{2}}+\int_{0}^{1 / t} \frac{d x}{1+x^{2}}\right)$$

Step-by-Step Solution

Answer

Answer: The derivative of the expression is $\frac{1}{1+t^2} + \frac{1}{t^4+t^2}$.

Step 1: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if F(x) is a function that is differentiable on the interval [a, b] and f(x) is the derivative of F(x), then: $$\frac{d}{dx}\int_a^b f(x) dx = f(x)$$ Our given expression has two integrals, so we need to apply the rule on both of them: $$\frac{d}{dt}\left(\int_0^t \frac{dx}{1+x^2}\right) + \frac{d}{dt}\left(\int_0^{1/t} \frac{dx}{1+x^2}\right)$$

Step 2: Apply the Chain Rule

Since both of our integrals have upper limits that are functions of t, we will need to use the chain rule when taking the derivative. The chain rule states that: $$\frac{d}{dt}[g(f(t))] = g'(f(t)) \cdot f'(t)$$ Now, let's apply the chain rule on the first integral: $$\frac{d}{dt}\left[\int_0^t \frac{dx}{1+x^2}\right] = \frac{1}{1+t^2} \cdot \frac{dt}{dt} = \frac{1}{1+t^2}$$ Applying the chain rule on the second integral: $$\frac{d}{dt}\left[\int_0^{1/t} \frac{dx}{1+x^2}\right] = \frac{1}{1+(1/t)^2} \cdot \frac{d(1/t)}{dt}$$

Step 3: Differentiate 1/t with respect to t

Now, we need to find the derivative of 1/t with respect to t. $$\frac{d(1/t)}{dt} = \frac{-1}{t^2}$$

Step 4: Substitute the derivative into the expression

Now that we have the derivative of 1/t, we can substitute it into the expression from Step 2: $$\frac{d}{dt}\left[\int_0^t \frac{dx}{1+x^2}\right]+\frac{d}{dt}\left[\int_0^{1/t} \frac{dx}{1+x^2}\right]=\frac{1}{1+t^2}-\frac{1}{t^2(1+(1/t)^2)}\cdot \frac{-1}{t^2}$$

Step 5: Simplify the expression

Finally, we need to simplify the expression by canceling out some terms and combining the fractions. $$\frac{1}{1+t^2}+\frac{1}{t^4+t^2}$$ So the final answer is: $$\frac{d}{dt}\left(\int_0^t \frac{dx}{1+x^2}+\int_0^{1/t} \frac{dx}{1+x^2}\right)=\frac{1}{1+t^2}+\frac{1}{t^4+t^2}$$

Key Concepts

Chain Rule

The Chain Rule is a fundamental principle in calculus that enables us to differentiate compositions of functions. When we have a function that is the result of another function operating within a third function, such as \( g(f(t)) \), the chain rule allows us to take the derivative of the outer function with respect to the inner function, and then multiply by the derivative of the inner function with respect to the original variable.

More formally, if \( g \) and \( f \) are functions, then the derivative of \( g(f(t)) \) with respect to \( t \) is expressed as \( g'(f(t)) \) multiplied by \( f'(t) \), or \( \frac{dg}{dt} = g'(f(t)) \cdot f'(t) \).

This rule is particularly helpful when dealing with integrals that have variable limits of integration, as seen in the given exercise. Without the Chain Rule, determining the derivative with respect to \( t \) of an integral from \( 0 \) to \( t \) or to \( 1/t \) would be significantly more challenging.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function’s value changes at a particular point. It's like taking a snapshot of a function at a certain moment and determining how fast it's moving at that instant.

The derivative of a function \( f \) with respect to a variable \( x \) is denoted by \( f'(x) \) or \( \frac{df}{dx} \). In the context of our exercise, differentiation is used to determine how the integral expressions change as the upper limit \( t \) changes. This is done by applying the Fundamental Theorem of Calculus, which links differentiation and integration.

Integration

Integration is the mathematical process of finding the integral of a function, which can be thought of as the reverse operation of differentiation. While differentiation tells us about the rate of change, integration tells us about the accumulation of quantities, such as area under a curve.

In our exercise, we're dealing with definite integrals, which are integrals with upper and lower limits. A definite integral of a function \( f(x) \) from \( a \) to \( b \) is typically written as \( \int_{a}^{b} f(x) \, dx \). It represents the total accumulation of \( f(x) \) across the interval from \( a \) to \( b \) on the \( x \) axis. Integration plays a key role in finding the area, volume, and in our exercise, the relationship between velocity and displacement.

Derivative

In calculus, the derivative of a function is a measure of how the function's output changes in response to changes in input. Graphically, for a function \( y = f(x) \), the derivative at a point corresponds to the slope of the tangent line to the function’s graph at that point. When the function is increasing, the derivative is positive, and when the function is decreasing, the derivative is negative.

The exercise focuses on finding the derivative of an integral, invoking the Fundamental Theorem of Calculus. This theorem states that if \( F \) is an antiderivative of function \( f \) over an interval, then the derivative of \( F \) will give us \( f \) again. In layman's terms, if you have an integrated function and you differentiate it, you end up with the original function—as long as the original function is continuous on the interval being considered.