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Gradients and First Principles: Exercise 2

by Batool Akmal
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    00:01 So, let?s look at the next exercise now, this one is a little bit easier because we?ve done an example like this before.

    00:07 So, let?s just do the algebra and hopefully enjoy this one.

    00:11 Remember the definition I?m gonna write that once again over and over again, so dy/dx as the limit of delta x tends to zero is f of x, plus delta x, minus f of x all over delta x.

    00:26 Okay, were gonna put that into the numbers or the numerical function.

    00:31 Remember what we?re doing each x function we are changing to x plus delta x and this in itself is just the original function which we?ll bring in, in a second.

    00:41 So, let?s write this all up, so we?ve got dy/dx as the limit of delta x tends to zero.

    00:47 We have 3 change the x squared to x, plus delta x squared, minus 2, change the x, to x plus delta x and then we have plus 7 at the end.

    00:59 Remember to take away the original function, so this is the unchanged function, so, we'll have 3x squared minus 2x, plus 7 and then it?s all being divided by delta x.

    01:11 Okay, here?s the common mistake that I see students make is this, so when we expand this, please be careful, remember to expand this as two bracket, so, this will going to give you a quadratic, so, if we just do that here, just to be extra cautious, we?ll leave as x plus delta x, x plus delta x first, we can expand of this a bit so, we?ve got 2x, minus 2 delta x, plus 7 and here once again be careful, because its multiplying through with the minus the signs inside will change, so you have minus 3x squared, plus 2x, minus 7 and it?s all being divided by delta x.

    01:48 Okay, expand this out firstly, so we?ve got 3, you're going to times it through, so we have x squared, plus x delta x, plus another x delta x, so that gives you 2x delta x, plus delta x squared, this is all fine, 2x minus 2 delta x, plus 7, minus 3x squared, plus 2x, minus 7 all over delta x, times it through, so we?ve got 3x squared, plus 6x delta x, plus 3 delta x squared, minus 2x, minus 2 delta x, plus 7, minus 3x squared, plus 2x, minus 7, we?re almost there, so all divided by delta x.

    02:35 Right, here you will see that we can cancel this function out, remember what I said, that if we?re doing it correctly this should usually cancel out, so you have positive 3x squared and a negative 3x squared.

    02:47 We have a negative 2x and a positive 2x and a minus 7 and a plus 7.

    02:53 If I do this in two steps, so we are now left with 6x, delta x, plus 3 delta x squared, minus 2 delta x, all over delta x. You now see that you?ve got a delta x common in each term here, here, here, and common with the denominator, so you can cancel delta x out, and cancel this one, with this one of them and this term.

    03:18 So, we're now left with 6x, plus 3 delta x, minus 2, and lastly don?t forget to apply this, we want our final answer of dy/dx as the limit of delta x tends to zero, that means that anything that had delta x in it, take it to zero, because it?s 3 times zero will give you zero, giving you a final answer of 6x, minus 2.

    03:43 So that is your differential, so we?ll also do this the faster and the shorter way, just because it?s easier and you'll see that you?ll prefer it.

    03:53 So if I just write this function out again so we can look at each term and what we?re doing to it minus 2x, plus 7.

    04:01 Remember what we discuss in the previous lecture that if you are differentiating this, you bring the power to the front and you minus 1 from this power.

    04:09 So you decreased the power by 1. So, let?s do that one bit at a time, so we?ve got 2 times, the 3 stays, that doesn?t bother us.

    04:15 We?ve got x to the power of 2 minus 1, so, you have to do this in order, bring the power to the front and decreased the power by 1.

    04:22 Again you do this for the next term, so the power of x now is just 1, so you can bring one to the front. We then times it with 2 and we have x to the 1, minus 1. And lastly, this constant here, will disappear and it will go to zero, and that?s because you can imagine writing this as 7x to the zero, so anything to the power of zero is just 1, so 7x to the zero is just 7, when you bring the power to the front, when you?re differentiating it, it?s just going to be zero multiplied by 7x, which is just zero.

    04:58 So your constants will always disappear, so any numbers that you see at the end of your functions that you?re differentiating, they just go away, so, you can just ignore them.

    05:07 Here we have 6x to the 1, when we tidy this up and then this gives you 2x to the zero is just 1, so like we've discussed before, anything to the power of zero is just 1.

    05:20 So this is your final answer, 6x, minus 2, which is exactly what we?ve got, the same value that we've got when we differentiated from first principles, but you can see, that thanks to the modern day mathematicians and modern calculus, things have become a lot easier for us, until we encouraged that you use this faster method.

    05:39 Unless, we actually need to go back to basics and derive and prove using first principles.


    About the Lecture

    The lecture Gradients and First Principles: Exercise 2 by Batool Akmal is from the course Calculus Methods: Gradients and First Principles.


    Author of lecture Gradients and First Principles: Exercise 2

     Batool Akmal

    Batool Akmal


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