91Ó°ÊÓ

The Leibniz rule is a technical but powerful tool for differentiation that extends the concept of differentiation under the integral sign. It is particularly useful when dealing with definite integrals where the limits of integration involve a variable. In our problem, this variable is the square root of x.When directly differentiating such integrals, the Leibniz rule offers a structured way to manage the differentiation process without evaluating the entire integral first. The rule states that if you have an integral of the form \[ \frac{d}{dx} \int_{a(x)}^{b(x)} f(x,t) \, dt = f(x,b(x)) \cdot b'(x) - f(x,a(x)) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,t) \, dt \]In our exercise, applying the Leibniz rule simplifies since our lower limit is a constant (0), and the function inside the integral doesn’t depend explicitly on the variable x. Therefore, our focus is on the upper limit \( \sqrt{x} \) which affects \( f(t) \): here, \( \cos(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) \) gives the direct derivative.

The chain rule is a fundamental principle used in calculus for finding the derivative of composite functions. It is vital when differentiating functions that are nested within one another, such as the sine of a square root, as encountered in our exercise.Suppose you have a function \( f(g(x)) \); the chain rule states:\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]In our case, when we differentiate \( \sin(\sqrt{x}) \) after evaluating the integral \( \int_{0}^{\sqrt{x}} \cos t \, dt \), we identify \( f(u) = \sin(u) \) and \( u = \sqrt{x} \). Differentiating produces \( \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \), perfectly following the chain rule formula.Always remember, whenever you see an inside function deeply wrapped around an outside one, the chain rule is your best friend.

Definite integrals help compute the net area under a curve within a specified interval. In our exercise, we aimed to find the derivative of a definite integral with upper limit dependent on \( x \).A definite integral \( \int_{a}^{b} f(t) \, dt \) results in a specific value, determined by evaluating the integrand's antiderivative from \( a \) to \( b \). For our function \( \cos t \), the antiderivative is \( \sin t \), resulting in:\[ \int_{0}^{\sqrt{x}} \cos t \, dt = \sin(\sqrt{x}) - \sin(0) \]This simplifies to \( \sin(\sqrt{x}) \) since \( \sin(0) = 0 \). The result shows why initial evaluation yields a clearer target for subsequent differentiation.

Differentiation under the integral sign is an advanced technique that allows differentiating integrals with variable boundaries without fully solving the integral. This method, famously used by Richard Feynman, facilitates handling integrals where the limits depend on the variable of differentiation.For example, in our exercise, we use this method in Step 3 to take the derivative without evaluating the integral. By applying this technique directly to \( \int_{0}^{\sqrt{x}} \cos t \, dt \), we find:\[ \frac{d}{dx} \int_{0}^{\sqrt{x}} \cos t \, dt = \cos(\sqrt{x}) \cdot \frac{d}{dx}(\sqrt{x}) \]This process highlights the straightforward nature of direct differentiation under the integral sign, especially when combined with understanding the integral's limits. It streamlines finding derivatives when evaluating the entire integral beforehand isn’t feasible.