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Differentiating Integrals Under mild continuity restrictions, it is true that if\(F(x)=\int_{a}^{b} g(t, x) d t\)then \(F^{\prime}(x)=\int_{a}^{b} g_{x}(t, x) d t .\) Using this fact and the Chain Rule, we can find the derivative of\(F(x)=\int_{a}^{f(x)} g(t, x) d t\)by letting\(G(u, x)=\int_{a}^{u} g(t, x) d t\)where \(u=f(x) .\) Find the derivatives of the functions.\(EF(x)=\int_{x^{2}}^{1} \sqrt{t^{3}+x^{2}} d t\)

Short Answer

Expert verified
F'(x) = -2x\sqrt{x^6 + x^2}

Step by step solution

01

Rewrite the Function

First, identify the given function and rewrite it clearly. The function is given as:\[F(x) = \int_{x^2}^{1} \sqrt{t^3 + x^2} \, dt\]Here, the integral is defined from the lower bound \(x^2\) to the upper bound 1.
02

Define G(u, x)

We want to differentiate \(F(x)\) by using the general form:\[G(u, x) = \int_{x^2}^{u} g(t, x) \, dt\]Given the function, define \(G(u, x)\) as:\[G(u, x) = \int_{x^2}^{f(x)} g(t, x) \, dt\]where \(u = f(x) = 1\), thus we actually have:\[G(1, x) = \int_{x^2}^{1} \sqrt{t^3 + x^2} \, dt\]
03

Apply the Leibniz Rule

To differentiate \(F(x)\), apply the Leibniz Rule, which is a suitable formula for differentiating integrals with variable limits:\[F'(x) = g(f(x), x) \cdot f'(x) - g(x^2, x) \cdot (x^2)'\]
04

Evaluate Each Term

Evaluate the two terms obtained from the application of the Leibniz Rule:1. For the first term: - Compute \( g(f(x), x) = \sqrt{1^3 + x^2} = \sqrt{1 + x^2} \) - Compute \( f'(x) = 0 \), as \( f(x) = 1 \) (a constant) - Therefore, the first term: \( \sqrt{1 + x^2} \cdot 0 = 0 \)2. For the second term: - Compute \( g(x^2, x) = \sqrt{(x^2)^3 + x^2} = \sqrt{x^6 + x^2} \) - Compute \( (x^2)' = 2x \) - Therefore, the second term: \( -\sqrt{x^6 + x^2} \cdot 2x \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Leibniz Rule
When solving problems involving integrals with variable limits, the Leibniz Rule is very helpful. It provides a method to differentiate an integral whose limits are functions of the variable of differentiation. In simple terms, if you have an integral such as \( F(x) = \int_{a(x)}^{b(x)} g(t, x) \, dt \), the derivative \( F'(x) \) can be found as:
  • Differentiate the upper limit \( b(x) \) to get \( b'(x) \), and then multiply it by the integrand evaluated at the upper limit: \( g(b(x), x) \).
  • Minus the derivative of the lower limit \( a(x) \) multiplied by the integrand evaluated at the lower limit: \( g(a(x), x) \) and the derivative of the lower limit \( a'(x) \).
This results in: \[F'(x) = g(b(x), x) \cdot b'(x) - g(a(x), x) \cdot a'(x)\] By applying this rule, you can easily tackle complex integrals that might seem daunting. In our example, this approach simplified the derivative of \( F(x) = \int_{x^2}^{1} \sqrt{t^3 + x^2} \, dt \), by considering the variable lower limit and treating the upper limit as a constant.
Chain Rule
The Chain Rule is fundamental when working with compositions of functions. In calculus, it's used to calculate the derivative of a composite function. When you have a situation where one quantity depends on another, which in turn depends on a third, the Chain Rule becomes essential.Simply put:
  • If you have \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
  • It signifies differentiating the outer function evaluated at the inner function and multiplying by the derivative of the inner function.
In the context of differentiating integrals, the Chain Rule helps connect the affect of changing limits of integration. Sometimes the formula of an integrand might depend on external variables, and the limits themselves might be functions of those variables. Combining the Chain Rule with other techniques like the Leibniz Rule often results in a more manageable approach for solving these types of calculus problems.
Variable Limits of Integration
Variable limits of integration occur when the limits of an integral are not constants. Instead, they are functions that depend on the variable of differentiation. This is what makes finding their derivative a bit more complex compared to ordinary definite integrals.Consider an integral with variable limits: \( F(x) = \int_{f(x)}^{g(x)} h(t, x) \, dt \).
  • The lower limit is \( f(x) \), and the upper limit is \( g(x) \).
  • To find the derivative, we use both the Chain Rule and the Leibniz Rule.
With these rules:
  • Upper limit contributes \( h(g(x), x) \cdot g'(x) \).
  • Lower limit contributes \( -h(f(x), x) \cdot f'(x) \).
Thus, differentiating integrals with variable limits lets us dynamically consider how the changes in bounds affect the integral's value. This is precisely why combining such concepts is crucial in effective calculation, especially with integrals like \( \int_{x^2}^{1} \sqrt{t^3 + x^2} \, dt \). By recognizing and applying these principles, handling complex calculus problems becomes a more straightforward task.

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Most popular questions from this chapter

The fifth-order partial derivative \(\partial^{5} f / \partial x^{2} \partial y^{3}\) is zero for each of the following functions. To show this as quickly as possible, which variable would you differentiate with respect to first: \(x\) or \(y ?\) Try to answer without writing anything down. $$ \begin{array}{l}{\text { a. } f(x, y)=y^{2} x^{4} e^{x}+2} \\ {\text { b. } f(x, y)=y^{2}+y\left(\sin x-x^{4}\right)} \\ {\text { c. } f(x, y)=x^{2}+5 x y+\sin x+7 e^{x}} \\ {\text { d. } f(x, y)=x e^{y^{2} / 2}}\end{array} $$

The condition \(\nabla f=\lambda \nabla g\) is not sufficient Although \(\nabla f=\lambda \nabla g\) is a necessary condition for the occurrence of an extreme value of \(f(x, y)\) subject to the conditions \(g(x, y)=0\) and \(\nabla g \neq 0,\) it does not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to find a maximum value of \(f(x, y)=x+y\) subject to the constraint that \(x y=16 .\) The method will identify the two points \((4,4)\) and \((-4,-4)\) as candidates for the location of extreme values. Yet the sum \((x+y)\) has no maximum value on the hyperbola \(x y=16\) . The farther you go from the origin on this hyperbola in the first quadrant, the larger the sum \(f(x, y)=x+y\) becomes.

Find all the second-order partial derivatives of the functions in Exercises \(41-50 .\) $$r(x, y)=\ln (x+y)$$

In Exercises \(49-52\) , find an equation for and sketch the graph of the level curve of the function \(f(x, y)\) that passes through the given point. $$ f(x, y)=\sqrt{x+y^{2}-3}, \quad(3,-1) $$

Let $$f(x, y)=\left\\{\begin{array}{ll}{y^{3},} & {y \geq 0} \\ {-y^{2},} & {y<0}\end{array}\right.$$ Find \(f_{x}, f_{y}, f_{x y},\) and \(f_{y x},\) and state the domain for each partial derivative.

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