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Leibniz's Rule Proof With Definition of a Derivative |
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| Sep2-12, 10:41 PM | #1 |
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Leibniz's Rule Proof With Definition of a Derivative
1. The problem statement, all variables and given/known data
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. 2. Relevant equations http://upload.wikimedia.org/math/4/2...c4115e7b5d.png 3. 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? |
| Sep2-12, 11:18 PM | #2 |
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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? |
| Sep3-12, 09:52 AM | #3 |
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