Showing the properties of differentiating an integral

  • Thread starter greswd
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  • #1
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How do we show that

[tex]\frac{d}{dt}\left[\int\!y\,\mathrm{d} x\right] = y\,\frac{dx}{dt}[/tex]
 

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  • #2
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How do we show that

[tex]\frac{d}{dt}\left[\int\!y\,\mathrm{d} x\right] = y\,\frac{dx}{dt}[/tex]
Is this a homework problem?
 
  • #3
HallsofIvy
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How do we show that

[tex]\frac{d}{dt}\left[\int\!y\,\mathrm{d} x\right] = y\,\frac{dx}{dt}[/tex]
Are we to assume, here, that y and x are functions of t? If we assume that y is a function of x only (with no "t" that is not in the "x") and x is a function of t, then we an write y(x(t)).

Of course, then [tex]F(x)= \int y dt[/tex] is the function such that dF/dx= y. Given that, we have that [itex]d/dt(\int y dx)= dF/dt= (dF/dx)(dx/dt)= y(x)(dx/dt)[/itex] by the chain rule.
 
  • #4
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Is this a homework problem?

Nope. Homework questions are usually standard, and answers are all in the textbooks.
I came up with this problem just out of curiosity.


Anyway, thanks for the solution HallsofIvy
 
Last edited:
  • #5
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Nope. Homework questions are usually standard, and answers are all in the textbooks.

I wish my textbooks had the answers!
 
  • #6
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I wish my textbooks had the answers!

not the exact answers, but they all follow the same template
 

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