Derivative by Leibniz's integral rule

In summary, when using Leibniz's integral rule, the partial derivative of τ with respect to x is equal to the integral of the partial derivative of R with respect to x plus the product of R(T) and the partial derivative of T with respect to x. However, the value of ∂R/∂x may or may not be zero, depending on the form of R(T), which is determined by the dummy variable θ. When finding the value of ∂τ/∂x at a specific point (x=c), it is equal to the product of R(T) evaluated at that point and the partial derivative of T with respect to x evaluated at the same point.
  • #1
Undoubtedly0
98
0
Hi all. I am looking at the functions [itex]\tau(x,y)[/itex], [itex]T(x,y)[/itex], and [itex]R(T)[/itex] related by

[tex] \tau(x,y) = \int_{\pi}^{T(x,y)} R(\theta) d\theta .[/tex]

It seems that by Leibniz's integral rule

[tex] \frac{\partial \tau}{\partial x} = \int_\pi^{T} \frac{\partial R}{\partial x} d\theta + R(T)\frac{\partial T}{\partial x}. [/tex]

It seems that [itex]\partial R / \partial x [/itex] need not be zero, yet another resource tells me that

[tex] \frac{\partial \tau}{\partial x} = R(T)\frac{\partial T}{\partial x} .[/tex]

Have I gone wrong somewhere? Thanks!
 
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  • #2
Sometimes ∂R/∂x is zero, sometimes not. Here you have written R(θ) which is often a hint that ∂R/∂x is zero, otherwise one usually writes R(θ,x).
 
  • #3
[itex]\theta[/itex] is a dummy variable that stands for [itex]T = T(x,y)[/itex], which I think means that [itex]R(\theta) = R(\theta(x,y))[/itex], correct?
 
  • #4
$$\dfrac{d}{dt}\int^{\mathrm{b}(t)}_{\mathrm{a}(t)} \mathrm{f}(x,t) \, \mathrm{dx}=\int^{\mathrm{b}(t)}_{\mathrm{a}(t)} \dfrac{\partial \mathrm{f}}{\partial t} \, \mathrm{dx}+\mathrm{f}(\mathrm{b}(t),t) \mathrm{b} ^ \prime (t)-\mathrm{f}(\mathrm{a}(t),t) \mathrm{a} ^ \prime (t)$$

No the dummy variable takes all the values in a set like (pi,T), it does not depend on the end point. The other term gives the effect of the boundary. Leibniz's integral rule breaks the dependence of the integral on the variable into dependence on the the interior and the boundary. Usual examples are cars on a highway or water in a pipe or electricity in a wire.
 
  • #5
So if we take it to be the case that

[tex] \frac{\partial \tau}{\partial x} = R(T)\frac{\partial T}{\partial x},[/tex]

then what would be the value of [itex] \partial \tau/\partial x |_{x=c} [/itex]?

[tex] \left.\frac{\partial \tau}{\partial x}\right|_{x=c} = R(T)|_{x=c} \cdot\left.\frac{\partial T}{\partial x}\right|_{x=c}[/tex]

What is meant by [itex] R(T)|_{x=c}[/itex]?
 
  • #6
Undoubtedly0 said:
What is meant by [itex] R(T)|_{x=c}[/itex]?

[tex]R(T)=R(T(x,y))[/tex]

Let's say you're given the point (a,b). Plug that point into T and it returns a number (call it c): T(a,b)=c. Then you plug this number into R to get R(T): R(T(a,b))=R(c)=d.

If you're only given x=a but y remains a variable, then T(a,y) is a function dependent only on the variable y. Thus R(T) is also a function dependent only on y. Plugging in a particular value of y will return a single number for R(T).
 

1. What is Leibniz's integral rule?

Leibniz's integral rule, also known as the Leibniz product rule or the product rule for integrals, is a formula used to find the derivative of a function that is expressed as the product of two other functions.

2. How is Leibniz's integral rule used in calculus?

Leibniz's integral rule is used in calculus to find the derivative of a function that cannot be easily differentiated using the basic rules of differentiation. It allows us to find the derivative of a product of functions by breaking it down into simpler parts.

3. What is the formula for Leibniz's integral rule?

The formula for Leibniz's integral rule is: d/dx ∫(u(x)v(x)) dx = u(x)v'(x) + v(x)u'(x). This means that the derivative of the integral of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function.

4. Can Leibniz's integral rule be applied to any product of functions?

Yes, Leibniz's integral rule can be applied to any product of functions, as long as both functions are continuous and differentiable.

5. Why is Leibniz's integral rule important in mathematics?

Leibniz's integral rule is important in mathematics because it allows us to find the derivative of a product of functions, which is necessary in many applications of calculus, such as optimization problems and curve sketching. It also helps to simplify complicated integrals and make them easier to solve.

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