Differentiation Method

00:00 So the next question that we're going to answer is what does differentiation actually do.

00:06 Why are we differentiating functions and why are we using these complicated and easy methods to find derivatives.

00:13 Differentiation helps you to find the gradient of a curve at a particular point.

00:18 You may think that's a bit strange because the gradient of a curve changes all the time.

00:22 But you can find the gradient of a particular point on the curve just by substituting your x and y values into that point.

00:31 It really is quite amazing how that works.

00:33 That when you differentiate, you get a general gradient of the curve.

00:37 And then you can look at any specific point by just substituting numbers into it.

00:41 We'll try this out with some examples shortly.

00:44 First of all we'll look at practicing A, differential question.

00:50 So we can do this really quickly now because I won't make you do this using first principles.

00:54 We're just going to make sure that we can differentiate and differentiate correctly.

00:59 And then we'll move on to substituting numbers in finding the gradient at more specific points.

01:05 And eventually we'll move on to finding gradients of tangents and lines and even equations of tangents of lines at those specific points.

01:12 So let's have a look at this question.

01:14 We will differentiate y = 3x2 + 5x - 1 and we'll do it the faster way.

01:20 So remember what we said that dy by dx.

01:24 You bring the power to the front.

01:27 So we'll have 2.

01:28 3x, 3 stays as it is.

01:32 Decrease the power of x by 1.

01:34 If I write it out as the whole thing.

01:36 You still bring the power of x down.

01:39 So the power of x now is 1.

01:41 And then you have 5x to the 1 - 1 which just becomes x to the 0.

01:47 And then the constant in the end will disappear because it has a power of x to the 0.

01:52 This than gives you an answer of 6x + 5.

01:57 and that's your gradient.

01:59 So really that's now made our lives a lot easier because we don't have to apply differentiation the first principles, we can just do it the faster way.

02:06 Remember now that this isn't just dy by dx.

02:10 This is actually telling us something.

02:13 It's telling us the gradient of this equation.

02:15 It's telling us how steep it is and at any different, any given point.

02:20 So say for example if this is a curve.

02:26 And the equation to that curve is y = to 3x2 + 5x - 1.

02:32 If we were to sketch this curve, you could factorize this.

02:38 So we've got 3x and x.

02:41 And then you'll have 1 and 1.

02:43 If you try this out, you will see that no matter what signs that you put in here and you won't quite be able to factorize it.

02:50 So they are other methods if you recall.

02:53 So previously if you remember that you would have done the quadratic formula which states the x = -b+-squared(b-4ac)/2a.

03:07 Your a is the number next to the x squared.

03:12 So this is your a.

03:13 Where b is the number next to the x and c is the number by itself.

03:17 So if ever you are not able to factorize to find points of intersections of a curve, remember there's always backup plans that we could use this equation to find our solutions.

03:30 So X is equal to -5, +- square root of 5 squared, -4 X 3, multiplied by -1.

03:38 This is all over 2 x 3, x is equals to -5 as before, and then in the root 5 squared gives us 25 and -4 x 3, -1, gives us +12.

03:50 And this is still all over 6.

03:52 So we end up with -5 +- root of 37 all over 6.

03:57 There's no obvious simplication factors here, so we can just leave it as it is.

04:01 Therefore, we have two answers, one positive and one negative, due to the +- before the root.

04:09 So if I sketch this, just giving over a rough idea.

04:14 So we got a positive and a negative answer.

04:16 And we've got a quadratic.

04:18 What we just found for ourselves is that the gradient of this curve at any given point, so here or here or here, And all you have to do is to find the actual gradient at that point is to substitute this x value into this equation.

04:35 So you could find the gradient at this point.

04:37 If you know this x value, you're just substituting here, or you could find the gradient at this point.

04:44 You know the x value you just substitute in here to find your specific gradient at that point.

04:50 Let's look at our second example now.

04:53 Now this is the only way that they can really make our lives difficult, is by giving us differential or equations to differentiate which are little bit more complicated.

05:02 So you have square roots or you have x values at the bottom as denominators.

05:06 And how do we tweak it and how do we make it more differentiation friendly for ourselves.

05:12 So let's look at this question and see if we can make it a little bit easier.

05:15 And just apply the straight forward concepts of differentiation.

05:19 So I'm just going to write this question again so that we can move things around.

05:22 So we've got y = 4 over x cubed, plus 2x to the power of 5 minus root x.

05:30 Now the first thing that we need to do in order to make our lives little bit easier, is we don't want this as denominator.

05:37 So we need to bring this up.

05:38 And secondly we don't want things like these.

05:40 So we don't want roots.

05:42 We want powers.

05:43 Because powers we can deal with and we know what to do with the power of something.

05:48 We can bring it down and we can decrease that by 1.

05:50 So before I differentiate I'm going to rewrite this.

05:53 As you know by the rule of indices or the rule of denominators being at the bottom, you are allowed to bring this up.

06:01 So you can bring this value up, as long as you change the sign of the power.

06:05 So we can rewrite this as 4x to the minus 3.

06:09 So this is one of the rules of the indices.

06:11 So we've got 2x to the power 5 which doesn't bother us.

06:14 And then we've got x to the power of a half.

06:17 So remember now I haven't differentiated it.

06:20 I've just rewritten it in order to make it easier for myself.

06:23 And this is perhaps the only way that these differentiation questions can become a little bite mathematically challenging.

06:29 But all you have to do is just look at it, simplify a little bit for yourselves.

06:34 Make it more differentiation friendly and then we just apply the same rule over and over again no matter what they are asking.

06:42 So here if I dy by dx and then I'm going to bring the power down so minus 3.

06:48 I've got 4x to the minus 3 of minus 1.

06:52 So decrease the power by 1.

06:54 You can bring 5 down.

06:55 The 2 is a constant which stays as it is.

06:58 And then you take 1 away from it.

07:00 And now we bring the half down and remember it doesn't matter what the power is.

07:04 No matter how complicated that is.

07:06 You just apply the same rule to it.

07:08 And then I've got x to the half minus 1.

07:12 And then I've basically done the difficult thing.

07:15 All I have to do is tidy it up and that will give me my final answer.

07:19 So this gives me -12 when I multiply these two numbers together.

07:23 I have x to the minus 4.

07:25 5 times 2 gives me 10.

07:27 x to the 4.

07:29 And then I have minus the half x to the minus the half.

07:34 And if you ever struggle with working something like this out, you can always just do it on the side because you are not quite using a calculator a half minus 1.

07:42 You want 2 to be your common denominator.

07:45 So you want both of these numbers to be above 2.

07:48 This changes, this doesn't change any.

07:51 This stays as it is.

07:52 And in order to increase this whole fraction by 2, you multiply the bottom, with the top by 2 as well.

07:58 So that gives you minus 2 to give you minus 1 at the top and 2 at the bottom.

08:03 So that's where it comes from.

08:04 So if fractions ever become difficult and you're not using a calculator, there's no shame in working out on the side and then putting it back into your answer.

08:12 Final few things that we could do so, we can make this x to the minus 4 go down.

08:18 So we can make it positive, so that becomes x to the 4.

08:21 Plus 10x to the 4.

08:22 And then we can bring this negative x to the minus the half down.

08:27 So that becomes joins the 2 at the bottom.

08:30 So that becomes the half.

08:32 And if you rewrite this as a square root, we can also do that.

08:35 And all you can do now is just change this last setup here if you wanted too.

08:39 So you could write this as square root of x.

08:42 Just the way that they did in the question.

08:44 So we have 12 over x to the 4 plus 10x to the 4 and then -1 over 2.

08:53 And instead of x to the half, you can write as root x just to show that you understand where it comes from.

08:59 So once again you are using a rule of indices to convert it back and make it look tidier.

09:05 So the gradient of this complicated function here is given by this.

09:11 I'm not going to try and sketch this out.

09:13 But remember that you can find the gradient at any particular point on this curve just by making substitutions for these x values here.

09:23 So if for example you are looking at the gradient at x equals to 2.

09:27 All you have to do is substitute 2 instead of the x's and you will get your gradient numerical value, be it positive or negative.

09:34 It can tell you a lot about what the curve is doing.