00:01
Okay, so our function here is s of x is equal to 1 over x plus 1.
00:05
So to find the derivative here, s prime of x using the definition, well, we want to basically the definition of a function of f of x, if we have f prime of x, is going to be equal to the limit as h goes to 0 of f of x plus h minus the function, so minus f of x all divided by h.
00:25
So let's go to first here, and let's find s of x plus h.
00:28
So s of x plus h, that's going to be equal to 1 over, well, x plus h plus 1.
00:38
So he's putting in an x plus h for x in our function.
00:42
And then the derivative here is going to be equal to the limit as h goes to zero of, well, s of x of x, that's going to be 1 over x plus h minus s of x.
00:58
Plus 1 minus 1 over x plus 1 so minus the function minus 1 over x plus 1 and then all divided by h so now we just find a common denominator here of x plus 1 times x plus h plus 1 and then we get the limit as h tends to 0 of while we end up here with a x plus 1 minus x plus h plus divided by x plus one times that common denominator of so x plus one times x plus one times x plus h plus one and then this is that all divided by h so we have here the limit as h goes to zero of let's see while we get an x plus one minus x minus h minus one everything cancels except for a minus h.
02:01
So we have a minus h over the x plus one times x plus h plus one...