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## Showing the properties of differentiating an integral

How do we show that

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

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 Quote by greswd How do we show that $$\frac{d}{dt}\left[\int\!y\,\mathrm{d} x\right] = y\,\frac{dx}{dt}$$
Is this a homework problem?

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 Quote by greswd How do we show that $$\frac{d}{dt}\left[\int\!y\,\mathrm{d} x\right] = y\,\frac{dx}{dt}$$
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 $$F(x)= \int y dt$$ is the function such that dF/dx= y. Given that, we have that $d/dt(\int y dx)= dF/dt= (dF/dx)(dx/dt)= y(x)(dx/dt)$ by the chain rule.

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## Showing the properties of differentiating an integral

 Quote by Mark44 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

 Quote by greswd Nope. Homework questions are usually standard, and answers are all in the textbooks.