Leibniz's Rule Proof With Definition of a Derivative

  • #1
PhysicsIzHard
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Homework Statement



Use the definition of the derivative to show that if G(x)=[itex]\int[/itex][itex]^{u(x)}_{a}[/itex]f(z)dz, then [itex]\frac{dG}{dx}[/itex]=f(u(x))[itex]\frac{du}{dx}[/itex]. This is called Leibniz's rule.

Also, by thinking of the value of an integral as the area under the curve of the integrand (and drawing a picture of that area), convince yourself that the following is true: lim[itex]\underline{x\rightarrow0}[/itex][itex]\int[/itex][itex]^{a+x}_{a}[/itex]f(z)dz=lim[itex]\underline{x\rightarrow0}[/itex]f(a)[itex]\int[/itex][itex]^{a+x}_{a}[/itex]dz. A relation like this will probably be useful in your solution to this problem.


Homework Equations



http://upload.wikimedia.org/math/4/2/c/42cf4f4861ae1266b13104c4115e7b5d.png

The Attempt at a Solution



I have tried to sub G(x) into the definition of the derivative equation but that gets me no where. Any ideas anyone on where to start this?
 

Answers and Replies

  • #2
jbunniii
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Consider introducing the function h(x) defined by

[tex]h(x) = \int_{a}^{x} f(z) dz[/tex]

Then [itex]G(x) = h(u(x))[/itex]. What happens if you use the chain rule to differentiate this?
 
  • #3
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Homework Statement



Use the definition of the derivative to show that if G(x)=[itex]\int[/itex][itex]^{u(x)}_{a}[/itex]f(z)dz, then [itex]\frac{dG}{dx}[/itex]=f(u(x))[itex]\frac{du}{dx}[/itex]. This is called Leibniz's rule.

Also, by thinking of the value of an integral as the area under the curve of the integrand (and drawing a picture of that area), convince yourself that the following is true: lim[itex]\underline{x\rightarrow0}[/itex][itex]\int[/itex][itex]^{a+x}_{a}[/itex]f(z)dz=lim[itex]\underline{x\rightarrow0}[/itex]f(a)[itex]\int[/itex][itex]^{a+x}_{a}[/itex]dz. A relation like this will probably be useful in your solution to this problem.


Homework Equations



http://upload.wikimedia.org/math/4/2/c/42cf4f4861ae1266b13104c4115e7b5d.png

The Attempt at a Solution



I have tried to sub G(x) into the definition of the derivative equation but that gets me no where. Any ideas anyone on where to start this?

Try writing the expression for G (x + [itex]\Delta x[/itex]) and subtracting the expression for G(x).
 

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