# Integration function homework question

1. Dec 25, 2007

### mathboy

Question:

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)?

Last edited: Dec 26, 2007
2. Dec 25, 2007

### 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).

Last edited: Dec 25, 2007
3. Dec 25, 2007

### 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 ......

Last edited: Dec 25, 2007
4. Dec 25, 2007

### Gib Z

5. Dec 25, 2007

### 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. ......

Last edited: Dec 25, 2007
6. Dec 25, 2007

### 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?

Last edited: Dec 25, 2007
7. Dec 26, 2007

### mathboy

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?