# Showing the properties of differentiating an integral

 P: 147 How do we show that $$\frac{d}{dt}\left[\int\!y\,\mathrm{d} x\right] = y\,\frac{dx}{dt}$$
Mentor
P: 21,400
 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?
Math
Emeritus
Thanks
PF Gold
P: 39,682
 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.

P: 147
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
P: 460
 Quote by greswd Nope. Homework questions are usually standard, and answers are all in the textbooks.