Hi Melody, the RHS comes from a standard result usually referred to as ' differentiation under the integral sign '.
For an indefinite integral the result takes the form
\(\displaystyle \frac{\partial}{\partial \alpha}\int f(\alpha,x)dx = \int\frac{\partial}{\partial \alpha}f(\alpha,x)dx .\)
It says that the order of the integral and the differentiation thingy are interchangeable, alpha is a parameter.
That's not always true, but it will be true for all of the functions you or I are ever going to meet.
As a simple example, since
\(\displaystyle \int e^{\alpha x} dx = \frac{1}{\alpha}e^{\alpha x}, \text{then}\\ \frac{\partial}{\partial \alpha} \int e^{\alpha x}dx=\int\frac{\partial}{\partial \alpha}e^{\alpha x} dx= \int xe^{\alpha x}dx =\frac{\partial}{\partial \alpha}(\frac{1}{\alpha}e^{\alpha x})= (\frac{x}{\alpha}-\frac{1}{\alpha^{2}})e^{\alpha x}.\)
(You can check the result by integrating by parts).
The result also works for a definite integral, providing that the limits of integration are constants.
If one or both of the limits of integration are functions of the parameter then the result takes the form
\(\displaystyle \frac{\partial}{\partial \alpha} \int^{b}_{a}f(\alpha,x)dx =\int^{b}_{a}\frac{\partial}{\partial \alpha}f(\alpha,x)dx+ \frac{db}{d \alpha}f(\alpha,b)-\frac{da}{d \alpha}f(\alpha,a).\)
For the example in the question, the parameter is x and the integration is wrt t, so we have (for the integral),
\(\displaystyle \frac{\partial}{\partial x}\int^{x}_{0}[f(t)]^{2}dt= \int^{x}_{0}\frac{\partial}{\partial x}[f(t)]^{2} dt +\frac{d}{d x}(x).[f(x)]^{2}-\frac{d}{dx}(0).[f(0)]^{2} \\ =[f(x)]^{2}.\)
Differentiate a function of t partially wrt x and you get zero, differentiate zero wrt x and you get zero so the first and third terms disappear.
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I hope that Christmas has gone well for you, and my very best wishes for the new year.
Tiggsy
