I wouldn't say that it was completely incorrect. A bit hard to understand maybe, but I got the gist of it... Not using capital F's for integrals didnt help. He did say:dextercioby said:Whozum,you're completely incorrect.
F(43) is a constant, and so its derivative is 0 and can be ommited.whozum said:Since the upper limit is a function of x, the chain rule is applied. f(43) is omitted since it is a constant.
It seems to me that [tex]\frac{d}{dx} \int_{u(x)}^{v(x)} f(x,t)dt[/tex] would be [tex]f(v(x),t)\frac{\partial v}{\partial x}-f(u(x),t)\frac{\partial u}{\partial x}[/tex] but that isn't the answer, is it?saltydog said:Tell you what Infirmus, check out Leibnitz rule and kindly solve the following:
[tex]\frac{d}{dx} \int_{u(x)}^{v(x)} f(x,t)dt[/tex]
You can do it I bet! Just follow the rules and do all the differentiations. The answer is impressive. And after that, just come up with some examples for the functions and do some real ones, say:
[tex]\frac{d}{dx} \int_{x^2}^{e^x} (x^2t+xtSin[x]) dt[/tex]
No Alex, that's not quite right. Remember, the integration is being done with respect to t and so at some point, you'll get a function in t that you then substitute v(x) and u(x) where ever t occurs. So if anything, it would include expressions such as:alexmcavoy@gmail.com said:It seems to me that [tex]\frac{d}{dx} \int_{u(x)}^{v(x)} f(x,t)dt[/tex] would be [tex]f(v(x),t)\frac{\partial v}{\partial x}-f(u(x),t)\frac{\partial u}{\partial x}[/tex] but that isn't the answer, is it?
Very good Alex. Suppose we have:alexmcavoy@gmail.com said:Right, I don't know why I did that. Anyways, I can understand that part, it's just that I can't get past why there is this:
[tex]\int_{u(x)}^{v(x)}\frac{\partial f}{\partial x}dt[/tex]
added to the answer.
I think this is the answer:
[tex]\int_{u(x)}^{v(x)}\frac{\partial f}{\partial x}dt+f(x,v(x))\frac{\partial v}{\partial x}-f(x,u(x))\frac{\partial u}{\partial x}[/tex]
according to MathWorld.
Any ideas?
Thanks,
--Alex
The final partial treats u and v as constants. Justifications are needed to switch the order of limits but its:alexmcavoy@gmail.com said:Alright, I guess I didn't think to write it out that way. So the part I was confused about comes from the [tex]\frac{dH}{dx}[/tex] at the end right?
Thanks again,
--Alex.
Sorry to bring it up again, but everything comes out correctly except for the sign when I use the chain rule. I come up with the same thing, but every term is added (just like the chain rule is written). Why is this?saltydog said:Very good Alex. Suppose we have:
[tex]H(u,v,x)=u^2+2uv+x[/tex]
with:
[tex]u=f(x);\quad v=g(x)[/tex]
Then:
[tex]\frac{dH}{dx}=\frac{\partial H}{\partial u} \frac{du}{dx}+\frac{\partial H}{\partial v} \frac{dv}{dx}+\frac{dH}{dx}[/tex]
So let:
[tex]H(u,v,x)=\int_{v(x)}^{u(x)} f(x,t) dt[/tex]
Hello Alex:alexmcavoy@gmail.com said:Sorry to bring it up again, but everything comes out correctly except for the sign when I use the chain rule. I come up with the same thing, but every term is added (just like the chain rule is written). Why is this?