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Magazine Chapter 5: Problem 76
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. \(h(x)=\int_{0}^{x^{2}} \sqrt{1+r^{3}} d r\)
Short Answer
Expert verified
The derivative of \( h(x) \) is \( h'(x) = 2x\sqrt{1+x^6} \).
Step by step solution
01
Understand the Fundamental Theorem of Calculus, Part 1
The Fundamental Theorem of Calculus, Part 1, tells us that if \( F(x) \) is the integral of \( f(t) \) from \( a \) to \( x \), i.e., \( F(x) = \int_{a}^{x} f(t) \, dt \), then the derivative \( F'(x) \) is equal to \( f(x) \). This applies when the upper limit of the integral is \( x \).
02
Identify the components of the function
In the given problem, the function \( h(x) = \int_{0}^{x^{2}} \sqrt{1+r^{3}} \, dr \) can be seen as an integral with variable upper limit \( x^{2} \). Here, \( f(r) = \sqrt{1+r^{3}} \).
03
Apply the Chain Rule
Since the upper limit of integration is a function of \( x \) (i.e., \( x^2 \) instead of \( x \)), we need to use the Chain Rule. If \( G(u(x)) \) is the integral of \( f(t) \) from \( a \) to \( u(x) \), then \( \frac{d}{dx}G(u(x)) = f(u(x)) \cdot u'(x) \) according to the Chain Rule.
04
Differentiate using the Chain Rule
Here, \( u(x) = x^2 \) and thus \( u'(x) = 2x \). By the Chain Rule, the derivative of \( h(x) = \int_{0}^{x^{2}} \sqrt{1+r^{3}} \, dr \) is given by evaluating \( \sqrt{1+(x^{2})^{3}} \cdot 2x \).
05
Simplify the derivative
Substituting \( x^{2} \) into \( \sqrt{1+r^{3}} \) gives \( \sqrt{1+(x^2)^{3}} = \sqrt{1+x^6} \). Hence, the derivative \( h'(x) = \sqrt{1+x^6} \cdot 2x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule Application in Derivative of Integrals
The Chain Rule is a fundamental concept in calculus. It helps us find the derivative of a function that is composed of other functions. In the case of integrals with a variable upper limit like in our original problem, the Chain Rule is crucial.
When dealing with an integral \( G(u(x)) = \int_{a}^{u(x)} f(t) \, dt \)with a variable upper limit of integration, we can't directly apply the Fundamental Theorem of Calculus as usual. Instead, since the upper limit \(u(x)\) is itself a function of \(x\), we need to account for this dependency.
When dealing with an integral \( G(u(x)) = \int_{a}^{u(x)} f(t) \, dt \)with a variable upper limit of integration, we can't directly apply the Fundamental Theorem of Calculus as usual. Instead, since the upper limit \(u(x)\) is itself a function of \(x\), we need to account for this dependency.
- First, differentiate \( G(u(x)) \) with respect to \(x\), while keeping in mind that the function \(u\) is also dependent on \(x\).
- According to the Chain Rule and Fundamental Theorem, the derivative will be \( f(u(x)) \times u'(x) \).
- In our example, this means substituting the upper limit \(x^2\) into \(f(r)\) and then multiplying by the derivative \(2x\) of \(x^2\).
Derivative Calculation Techniques
Derivative calculation is a primary skill in calculus, used for finding the rate of change. Derivatives provide a functional slope at any given point and convey significant information about the function's behavior.
In standard practice, to find the derivative of an integral with a variable limit:
In standard practice, to find the derivative of an integral with a variable limit:
- Identify the integrand function \( f(r) \) within the integral limits.
- Recognize the variable limit, typically expressed as a function of \(x\), such as \(x^2\).
- Use the Chain Rule in coordination with the Fundamental Theorem of Calculus for differentiation.
- Evaluate the function \( f \) at the upper limit \( u(x) \) and multiply by the derivative of \( u(x) \).
Understanding Integrals with Variable Limits
An integral with a variable upper limit, such as \( h(x) = \int_{0}^{x^{2}} \sqrt{1+r^{3}} dr \),offers a unique challenge when differentiating.
Usually, integrals have a fixed upper and lower limit. However, when the upper limit is a function of \(x\), different rules apply. To differentiate such an integral, the roles of the Fundamental Theorem of Calculus and the Chain Rule are pivotal.
Usually, integrals have a fixed upper and lower limit. However, when the upper limit is a function of \(x\), different rules apply. To differentiate such an integral, the roles of the Fundamental Theorem of Calculus and the Chain Rule are pivotal.
- The Fundamental Theorem provides a foundational understanding, acting as a bridge between differentiation and integration theories. It helps translate an integral into a function that can be differentiated.
- With a variable limit, the theorem needs extending through the Chain Rule to manage the additional calculation complexity.
