Integration function homework question

[mathboy]

Question:
http://img167.imageshack.us/img167/8606/questionan4.jpg


Typing correction: g maps to R^n, and id_R should be id_R^n
Here id_R^n means the identity function on R^n. So the condition is saying g(x,b(x))=x.

Any ideas how to do this one? Should I begin with the substitution y=g(x,t)?

 

[Gib Z]

I don't really know what an identity map on R actually is, but otherwise I can get it down to $$f(x) = f( g(x, \beta (x))$$.

EDIT: Am I right in assuming that it is sort of like the identity function h(x) = x, but for only the first variable? Basically I'm asking, if that condition the same as; $$g(x,\beta (a) ) = x$$ as long as $$x,a \in \mathbb_{R}$$. If so, then my answer reduces down to f(x).

 

[mathboy]

I've retyped the condition g above.

So if the substitution y=g(x,t) is made, then when t = b(x), we have
y=g(x,b(x))=x by the condition on g.

Now f(g(x,t))=f(y) is now a function of y only, but y itself is a function of n+1 variables x_1,...,x_n, t. So now the chain rule when we take df(y)/dt will be ...

 

[Gib Z]

I used the differentiation under the integral method, this link should get any rigor questions out of the way: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

Now, wrt t the bounds are constants, so can you see how that integral evaluates to $$f(g(x,\beta (x)) - f(g(x,\alpha (x))$$?

 

[mathboy]

But Leibnit'z rule involves taking the derivative wrt, say, x, with the integral is wrt, say, t. That's not what's happening here in this question. The integral is wrt to t and the partial derivative also wrt to t. Leibnitz's rule does not apply here.

Also, I don't think we can just cancel out the two dt 's in the integrand because they are not the same dt 's. I think more rigour is required to PROVE the result. I'm attempting to apply the chain rule here:

f(g(x,t))= (fog)(x,t) so by the chain rule, (fog)'(x,t) = f'(g(x,t)*g'(x,t) yields a 1x(n+1) matrix and we are interested in the component that gives df(g(x,t))/dt only. ...

 

[mathboy]

Using the chain rule, I get df(g(x,t))/dt to be a messy sum of n terms.

I need to know: Is [Df(g(x,t))/Dt]dt = df(g(x,t)), if the first D is partial derivative and the second d ordinary derivative ? If so, how do I prove that?

 

[mathboy]

Is [Df(g(x,t))/Dt]dt = df(g(x,t)), if the first D is partial derivative and the second d ordinary derivative ? If so, how do I prove that?

I believe the answer is yes because in this case [Df(g(x,t))/Dt]dt is the integrand, and we are integrating with respect to t only. Thus we can view x as a constant (within the integral) and thus the partial derivative D/Dt can be treated as an ordinary derivative d/dt. Am I right?