Question

Let \(F(t)=\int_{g(t)}^{h(t)} f(u) d u\), where \(f\) is continuous and \(g\) and \(h\) are differentiable. Show that $$ F^{\prime}(t)=f(h(t)) h^{\prime}(t)-f(g(t)) g^{\prime}(t) $$ and use this result to find \(F^{\prime}(\sqrt{2})\), where $$ F(t)=\int_{\sin \sqrt{2} \pi t}^{t^{2}} \sqrt{9+u^{4}} d u $$

Short answer

\( F^{\prime}(\sqrt{2}) = 8\sqrt{2} - 3\sqrt{2}\pi \).

Step-by-step solution

Apply Leibniz Rule for Differentiation Under the Integral Sign

The expression for the derivative of an integral with variable limits, according to Leibniz's Rule, is given by:\[ F^{\prime}(t) = f(h(t)) h^{\prime}(t) - f(g(t)) g^{\prime}(t) \]where \( f(u) \) is continuous and \( g(t), h(t) \) are differentiable functions.

Identify Functions Inside the Integral

For the given function \( F(t) = \int_{\sin \sqrt{2} \pi t}^{t^2} \sqrt{9+u^4} \, du \), we identify:- \( g(t) = \sin \sqrt{2} \pi t \)- \( h(t) = t^2 \)- \( f(u) = \sqrt{9+u^4} \)

Differentiate the Upper Limit Function

Differentiate \( h(t) = t^2 \) to find \( h^{\prime}(t) = 2t \).

Differentiate the Lower Limit Function

Differentiate \( g(t) = \sin \sqrt{2} \pi t \) to obtain:\[ g^{\prime}(t) = \sqrt{2} \pi \cos \sqrt{2} \pi t \]

Apply Leibniz Rule Using Identified Functions

Substitute the derivatives and functions from Steps 2-4 into the Leibniz Rule formula:\[ F^{\prime}(t) = \sqrt{9 + (t^2)^4} \cdot 2t - \sqrt{9 + (\sin \sqrt{2} \pi t)^4} \cdot \sqrt{2} \pi \cos \sqrt{2} \pi t \]

Find the Derivative at Specific Point

To find \( F^{\prime}(\sqrt{2}) \), substitute \( t = \sqrt{2} \) into the expression: 1. For \( 2t \) term: \( 2\sqrt{2} \)2. For \( t^2 \) term inside \(f\): \( (\sqrt{2})^2 = 2 \)3. For \( \sin \sqrt{2} \pi t \): calculate \( \sin (2\pi) = 0 \)Thus, \[ F^{\prime}(\sqrt{2}) = \sqrt{9 + 2^4} \cdot 2\sqrt{2} - \sqrt{9 + 0^4} \cdot \sqrt{2} \pi \cdot 1 \]

Key concepts

Differentiation under the integral sign

Differentiation under the integral sign is a powerful technique, often associated with Leibniz's Rule, which facilitates taking the derivative of an integral where the limits of integration are themselves functions of the variable differentiating with respect to. This concept is especially significant when dealing with integrals in which the limits are not constants.The rule is expressed in the formula:\[F^{\prime}(t) = f(h(t)) h^{\prime}(t) - f(g(t)) g^{\prime}(t)\]Here, the integral transforms as each limit has its derivative considered and adjusted appropriately. This allows us to express how changes in the variable \(t\) affect the value of \(F(t)\). Remember these steps whenever you encounter a calculus problem demanding differentiation under the integral sign:

• Identify the functions in the integral and their limits.
• Differentiate these limits with respect to the intended variable.
• Apply them to Leibniz's formula to obtain the result.

This approach is highly applicable in various fields such as physics and engineering where dynamic changes are modeled mathematically.

Variable limits of integration

Variable limits of integration add complexity to finding derivatives, but they also allow for more flexible and natural expressions of problems, particularly in calculus. When the limits of an integral are not fixed constants but rather functions of another variable, handling them properly is essential.In our specific scenario:

• The lower limit is \( g(t) = \sin \sqrt{2} \pi t \).
• The upper limit is \( h(t) = t^2 \).

When calculating \( F^{\prime}(t) \), differentiating these variable limits correctly is crucial. The derivative of \( h(t) \) when applied in the Leibniz Rule contributes a term involving \( f(h(t)) h^{\prime}(t) \), and similarly for \( g(t) \), the term involves \( -f(g(t)) g^{\prime}(t) \).To handle variable limits, keep these tips in mind:- Differentiate each limit accurately.- Substitute the derivatives into the Leibniz Rule.- Maintain careful attention to minus signs, as they denote the direction of integration.

Calculus problems

Calculus is a branch of mathematics concerned with continuous change, and problems involving it often require a deep understanding of both integration and differentiation. Textbook exercises involving these principles demand students to apply theoretical knowledge in practical contexts such as the problem we're examining here. Key components of tackling these calculus problems effectively include:

• Understanding the fundamental theorems of calculus, including the Leibniz Rule.
• Being able to manipulate and move between different representations, like integral and differential forms.
• Noting the context, specifically how the limits of integration interact with the integrand.

Our exercise exemplifies these concepts and exercises your skills in differential calculus, emphasizing flexibility in mathematical thinking and attentive application of rules. Solving such problems builds a solid foundation for more advanced topics in calculus.

Continuous and differentiable functions

For many mathematical concepts to hold, such as those in calculus, it's important that functions are both continuous and differentiable. Continuous functions have no breaks, jumps, or points of discontinuity, making them smooth and predictable to work with, while differentiable functions have well-defined derivatives.In the context of our exercise:- The integrand \( f(u) = \sqrt{9+u^4} \) remains continuous everywhere due to the nature of the square root and polynomial functions.- The functions \( g(t) \) and \( h(t) \) are differentiable, meaning we can find their derivatives \( g^{\prime}(t) \) and \( h^{\prime}(t) \).These properties ensure the application of the Leibniz Rule is valid. When functions are both continuous and differentiable, it allows for ease in calculating derivatives and applying fundamental rules of calculus, making problem-solving straightforward.