Derivative of Definite Integral Conundrum

[ObjectivelyRational]

Homework Statement

The normal approach using the fundamental theorem of calculus seems inapplicable. I define a function B(R) based on a definite integral with one of the limits being R. One factor in the definite integral has R in it and that function vanishes to 0 at x = R.

Using the fundamental theorem I run into the problem that the derivative of B(R) evaluates to 0.

Homework Equations

K is just a constant greater than R.

The Attempt at a Solution

Reversing the sign and the limits of integration is as far as I got. If I do a straight replacement of x with R, Cos^-1(R/x) goes to Cos^-1(1) which is zero...

I'm trying to isolate G but this has me stumped.

 

[andrewkirk]

Try writing the two occurrences of $R$ in the formula as separate variables $u$ and $v$ and write $B(R)$ as a function of the two variables, each of which is a function of $R$, viz: $u(R)=V(R)=R$.

If you can do that then you can then use the http://tornado.sfsu.edu/Geosciences/classes/m430/TotalDerivative/Total_derivative.html formula to find $\frac{dB}{dR}$.

 

[Ray Vickson]

Homework Statement

The normal approach using the fundamental theorem of calculus seems inapplicable. I define a function B(R) based on a definite integral with one of the limits being R. One factor in the definite integral has R in it and that function vanishes to 0 at x = R.

Using the fundamental theorem I run into the problem that the derivative of B(R) evaluates to 0.

Homework Equations

View attachment 105677K is just a constant greater than R.

The Attempt at a Solution

Reversing the sign and the limits of integration is as far as I got. If I do a straight replacement of x with R, Cos^-1(R/x) goes to Cos^-1(1) which is zero...

I'm trying to isolate G but this has me stumped.

Have you forgotten (or perhaps, never learned) Leibniz' (Integral) Rule? See, eg.,
http://mathworld.wolfram.com/LeibnizIntegralRule.html