Chapter 5: Problem 11
\(5-14=\) Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. $$ y=\int_{0}^{\tan x} \sqrt{t+\sqrt{t}} d t $$
Short Answer
Expert verified
\( y'(x) = \sqrt{\tan x + \sqrt{\tan x}} \cdot \sec^2 x \)
Step by step solution
01
Identify the integral function
The given function is \( y = \int_{0}^{\tan x} \sqrt{t+\sqrt{t}} \, dt \). This is a definite integral where the upper limit is a function of \( x \).
02
Apply Part 1 of the Fundamental Theorem of Calculus
Part 1 of the Fundamental Theorem of Calculus allows us to differentiate an integral with a variable upper limit. It states that \( \frac{d}{dx} \left( \int_{a}^{u(x)} f(t) \, dt \right) = f(u(x)) \cdot u'(x) \).
03
Identify the components for differentiation
In our problem, \( a = 0 \), \( u(x) = \tan x \), and \( f(t) = \sqrt{t+\sqrt{t}} \). We need to find \( f(u(x)) \) and \( u'(x) \).
04
Substitute and find the derivative
First, substitute \( u(x) = \tan x \) into \( f(t) \):\[ f(u(x)) = \sqrt{\tan x + \sqrt{\tan x}} \]Next, find the derivative of \( u(x) \):\[ u'(x) = \sec^2 x \]
05
Multiply components to find \( y'(x) \)
Using Part 1 of the FTC, substitute the results from Step 4: \[ y'(x) = \sqrt{\tan x + \sqrt{\tan x}} \cdot \sec^2 x \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus, which concerns finding the rate at which a function changes. It's the process of calculating the derivative of a function, which tells us how the function value changes as its input changes. When dealing with integrals, especially those with a variable upper limit, differentiation plays a crucial role in understanding how the integral function changes. Differentiation with respect to a variable upper limit is simplified using the Fundamental Theorem of Calculus (FTC). This theorem connects differentiation and integration, enabling us to differentiate integrals where the upper boundary is a function rather than a constant.
- Example: For a function defined as an integral like \( y(x) = \int_{0}^{ an x} f(t) \), we see that the upper limit \( \tan x \) depends on \( x \).
- Calculation: The FTC allows finding the derivative of such a function through the chain rule in calculus.
Definite Integral
A definite integral computes the accumulated sum of a function within specific bounds, providing a precise area under a curve from one point to another. In our problem, the function is defined as a definite integral \( \int_{0}^{\tan x} \sqrt{t + \sqrt{t}} \, dt \). Here, the integral sums up the function \( \sqrt{t + \sqrt{t}} \) from the lower limit 0 to \( \tan x \), which is a variable based on \( x \).
- Visual Understanding: Graphically, a definite integral can be visualized as the area between the curve, the \( t \)-axis, and the vertical lines at the bounds of the integration.
- Outcome: The value of this integral changes depending on the upper limit \( \tan x \), making each evaluation unique to the specific argument \( x \).
Variable Upper Limit
The concept of a variable upper limit involves integrals whose upper boundary is not a constant but rather a function of another variable, like \( x \) in the exercise. This turns the integral into a function of \( x \), introducing dynamic behavior based on how the variable influences the limit.When we have \( \int_{0}^{u(x)} f(t) \, dt \), the integral's value depends on the function \( u(x) \) for the upper limit. Changes to \( x \) thus directly affect the overall evaluation of the integral.
- Example: In our problem, \( u(x) = \tan x \), illustrating how the boundary varies as \( x \) changes.
- Application: Using the fundamental theorem effectively requires calculating both the function evaluated at the upper limit, \( f(u(x)) \), and the derivative \( u'(x) \).