Gradients and First Principles: Exercise 3

00:00 So the last exercise that we have is differentiating f of x equals to x to the n.

00:08 Now this is interesting because you may remember that we use this function to define what differentiation is. So we said that the differential of this is dy/dx and we use this as our basic method for the faster method really is we bring the power to the front and we decrease the power by 1.

00:26 So that is the faster way really, we've almost done that before but the idea is that we wanted now get this result and we need to prove it by first principles so if we apply first principles to it, it's almost like we're doing a mathematical proof to show that x to the n will always differentiate to nx to the n minus 1.

00:45 So let's apply differentiation from first principles and actually produce our proof here.

00:52 So we're going to say dy/dx as the limit of delta x tends to zero.

00:57 And we're going to use our equation, f of x plus delta x minus f of x divided by delta x.

01:07 Let's just put our function in so rather than x now, we're going to have x plus delta x to the power of n, subtract the original function and divide it by delta x.

01:20 Okay, once again we're faced by this type of an expansion which we don't like but remember what we said, we can tackle this problem, using binomial expansion.

01:31 So what we could do again is use binomial expansion on this term here, substitute it back into this function and hopefully reach this result which we're aiming to do, today.

01:43 So let's put this here, so let's just use binomial theorem, firstly.

01:49 Remember what we said with binomial theorem, you start with the highest power of x, so we have x to the end multiplied by delta x to the power of zero, next term, or you can say this is over zero factorial to be more precise.

02:03 We then bring the power down and we have x to the power of n minus 1 and then we have delta x to the power of 1 over 1 factorial.

02:12 There are more terms but if you recall what I said earlier that if you have any term with delta x squared, one of the delta x cancels out but the others goes to zero. So again a little bit of forward thinking in mathematics makes life a lot easier so we'll just leave it out that.

02:30 We now simplify this a little bit so that x to the n, anything to the power of zero is just 1 so we can ignore that.

02:37 Here we have nx to the power of n minus 1 multiplied by delta x and one factorial is just one, and there are other terms but we're not including them because we know they will cancel out. So that's my expansion for this bracket here, we now put this back into our differentiation from first principles, so dy/dx as the limit of delta x tends to zero. If we put that expansion n, we have n, x to the power of n minus 1 delta x.

03:12 Leave that for now, ignore the rest of the terms, minus x to the power of n, all over delta x.

03:19 You will see that when you bring this out of the brackets, x, n minus 1, multiplied by delta x, minus x to the n over delta x as we predict most often that these two terms will cancel out, you also have the delta x and the delta x canceling out, leaving us with our result n, x, n minus 1 we're there but just to be precise let's just say dy/dx as the limit of delta x tends to zero is n, x, n minus 1. And again this would apply if I made the effort to write all of these terms here but just to save time I haven't. And so there is nothing that I need to take to zero because I already know that all of these terms will equal to zero because they will all include a delta its term left over.

04:11 So my final result, you'll see matches with my first result which is what we've said, so we've in fact proved not only doing differentiation of the first principles but also differentiating the faster way but by an algebraic and mathematical proof.