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Chapter 14: Problem 51

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 in Exercises 51 and \(52 .\) \(F(x)=\int_{0}^{x^{2}} \sqrt{t^{4}+x^{3}} d t\)

Short Answer

Expert verified
The derivative \( F'(x) \) is \( 2x \sqrt{x^8 + x^3} + \int_{0}^{x^2} \frac{3}{2}x^2(t^4 + x^3)^{-1/2} \, dt \).

Step by step solution

01

Define the Function G(u, x)

First, we define function \( G(u, x) \) given by the integral \( G(u, x) = \int_{0}^{u} \sqrt{t^4 + x^3} \, dt \). This is to help transform the original problem using the variable \( u = f(x) = x^2 \).
02

Differentiating G(u, x) with Respect to x

Next, differentiate \( G(u, x) \) with respect to \( x \) using the given rule \( \frac{\partial}{\partial x} G(u, x) = \int_{0}^{u} g_{x}(t, x) \, dt \). Here, \( g(t, x) = \sqrt{t^4 + x^3} \), so we find the partial derivative of \( g \) with respect to \( x \): \( g_{x}(t, x) = \frac{3}{2}x^{2}(t^4 + x^3)^{-1/2} \).
03

Applying the Chain Rule

Now, find \( F'(x) \) using both the chain rule and the fundamental theorem of calculus as follows:- Differentiate the outer function \( G(u, x) \) using the chain rule, respecting the dependency \( u = x^2 \):\[ F'(x) = \frac{\partial G}{\partial u}(x^2, x) \cdot \frac{d}{dx}(x^2) + \int_{0}^{x^2} \left(\frac{3}{2}x^{2}(t^4 + x^3)^{-1/2}\right) \, dt \]- The partial derivative with respect to \( u \) gives:\( \frac{\partial G}{\partial u}(x^2, x) = \sqrt{(x^2)^4 + x^3} = \sqrt{x^8 + x^3} \).
04

Calculating F'(x) Fully

Putting it all together, substitute \( \frac{\partial G}{\partial u}(x^2, x) \) into the previous result, then calculate:\[ F'(x) = \sqrt{x^8 + x^3} \cdot (2x) + \int_{0}^{x^2} \frac{3}{2}x^{2}(t^4 + x^3)^{-1/2} \, dt \]The first term simplifies directly, while the second integral term must be evaluated separately using specific conditions or additional techniques as specified in problems, typically focusing only on the first part for typical comparisons.

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Key Concepts

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

Chain Rule
The Chain Rule is a fundamental technique in calculus, particularly useful when dealing with composite functions. If you have a composite function like \( F(x) = G(f(x)) \), the Chain Rule states that the derivative \( F'(x) \) is the product of the derivative of \( G \) evaluated at \( f(x) \) and the derivative of \( f(x) \). This is written as:
  • \( F'(x) = G'(f(x)) \, f'(x) \)
When differentiating integrals, the Chain Rule helps manage functions that have limits dependent on another variable. In our problem, \( u = x^2 \) serves as an upper limit for the integral, treated as a function of \( x \). Here, we apply the Chain Rule as part of differentiating \( G(u, x) \). This involves differentiating \( G \) partially with respect to \( u \), and then multiplying by the derivative of the upper limit function \( u = x^2 \), which is \( 2x \).
The Chain Rule is crucial in simplifying and finding derivatives even with seemingly complex nested functions.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links differentiation and integration, two core concepts in calculus. It is divided into two parts. The first part deals with the evaluation of definite integrals, while the second part allows us to differentiate under the integral sign.
In simple terms, if \( F(x) \) is the integral of a function \( g(t) \), the theorem states that the derivative of \( F(x) \) with respect to \( x \) gives us back the original integrand function. Specifically,
  • \( rac{d}{dx} \int_{a}^{b} g(t, x) \, dt = g(b, x) \, rac{db}{dx} - g(a, x) \, rac{da}{dx} + \int_{a}^{b} g_x(t, x) \, dt \)
In our example, since \( a \) is a constant zero and \( b = x^2 \), only the derivative with respect to the upper limit involves change, showcasing how the Fundamental Theorem fits into differentiating an integral with variable limits.
Partial Derivatives
Partial derivatives are used when dealing with multivariable functions. They describe how a function changes as one of several variables changes, keeping the others constant.
In the context of the integral \( F(x) = \int_{0}^{x^2} \sqrt{t^4 + x^3} \ dt \), the integrand function \( g(t, x) = \sqrt{t^4 + x^3} \) depends on both \( t \) and \( x \). To apply differentiation rules, we find the partial derivative with respect to \( x \), denoted \( g_x(t, x) \). This gives insight into the behavior of the function as \( x \) changes, holding \( t \) constant.
For example, the partial derivative \( g_x(t, x) = \frac{3}{2}x^{2}(t^4 + x^3)^{-1/2} \) shows how sensitive the squared term in \( x \) affects \( g(t, x) \), thereby influencing the integral's overall differentiation outcome when \( x \) varies.

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Most popular questions from this chapter

Two dependent variables Find \(\partial x / \partial u\) and \(\partial y / \partial u\) if the equations \(u=x^{2}-y^{2}\) and \(v=x^{2}-y\) define \(x\) and \(y\) as functions of the independent variables \(u\) and \(v,\) and the partial derivatives exist. (See the hint in Exercise \(69 . )\) Then let \(s=x^{2}+y^{2}\) and find \(\partial s / \partial u .\)

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x y+y z\) subject to the constraints \(x^{2}+y^{2}-\) \(2=0\) and \(x^{2}+z^{2}-2=0.\)

Establish the fact, widely used in hydrodynamics, that if \(f(x, y, z)=0,\) then \begin{equation}\left(\frac{\partial x}{\partial y}\right)_{z}\left(\frac{\partial y}{\partial z}\right)_{x}\left(\frac{\partial z}{\partial x}\right)_{y}=-1.\end{equation} (Hint: Express all the derivatives in terms of the formal partial derivatives \(\partial f / \partial x, \partial f / \partial y,\) and \(\partial f / \partial z . )\)

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Determine the distance from the line \(y=x+1\) to the parabola \(y^{2}=x .\) (Hint: Let \((x, y)\) be a point on the line and \((w, z)\) a point on the parabola. You want to minimize \((x-w)^{2}+(y-z)^{2}.)\)

In Exercises \(53-60,\) sketch a typical level surface for the function. $$ f(x, y, z)=x+z $$
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