0:00:02.620,0:00:06.270
In previous tutorials, you've[br]differentiated from first

0:00:06.270,0:00:12.736
principles functions such as[br]c, x, 2x, x to the power n,

0:00:12.736,0:00:19.604
where n is any power, sine x,[br]cosine x, e to the power x

0:00:19.604,0:00:22.446
and log to the base e of x.

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Now what we're going to do in[br]this tutorial is to construct

0:00:28.016,0:00:31.238
a table of these standard[br]derivatives, and when we've done

0:00:31.238,0:00:34.460
that, we'll look at some[br]variations of those standard

0:00:34.460,0:00:36.966
derivatives and add those to[br]our table.

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So let's start with a function.

0:00:43.650,0:00:44.730
f of x.

0:00:46.240,0:00:47.650
And add derivative.

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Which may be referred to as,[br]df by dx or f dashed of x.

0:01:01.570,0:01:07.352
So let's start with c, well c is a[br]constant. It's a straight line

0:01:07.352,0:01:11.069
and it's horizontal. So I[br]gradient function is 0.

0:01:12.980,0:01:16.956
x, the derivative of x is 1.

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2x, the derivative is 2.

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If we have x to the power n.

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Then the derivative is n times x[br]to the power of n - 1,

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where n is a real number.

0:01:38.860,0:01:43.276
sine x, the derivative is cos x.

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cos x, the derivative is -sine x.

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e to the power x, that's the one that[br]stays the same, e to the power x.

0:01:58.830,0:02:01.846
And log to the base e of x.

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Where the derivative is 1 over x.

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Let's look now at a variation[br]of some of these, we'll come

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back to the table in just a minute.

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OK, what if we have f of x[br]equals, instead of just sine x,

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we have 2 sine x.

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Well, this is just the same as

0:02:32.130,0:02:35.085
taking sine x and taking two of them.

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So when we find the derivative,[br]we actually find the

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derivative of sine x and take[br]two of those.

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Well, the derivative of sine x[br]is cos x.

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So the derivative of 2 sine x[br]is 2 cos x.

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Similarly, if we take f of x is[br]-5 sine x, then our derivative

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f dashed of x is -5 cos x.

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And this in fact is the case for

0:03:10.758,0:03:15.520
any constant multiplied by the function.

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So if we have f of x equals[br]C times sine x.

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Then f dashed of x is going to[br]be C times cos x.

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And this is called the[br]constant multiplier rule.

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So if we have...

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C times our function of x.

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And we want to differentiate it.[br]So we want to find the

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derivative with respect to x.

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It's actually just C times the[br]derivative df by dx or

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C times our f dashed of x.

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Now let's prove this from[br]first principles.

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If we take our function of x[br]to be g of x and

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it's equal to some constant times f of x.

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Now, our definition of our derivative

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is g dashed x equals the limit [br]as delta x approaches 0 of

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g of x plus delta x, minus g of x,[br]all divided by delta x.

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Which equals, again our limit[br]as delta x approaches 0.

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And in this case our g of x plus delta x [br]is C of f of x plus delta x.

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So we substitute that in.

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Minus our g of x which is C f of x.

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All divided by delta x.

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And here I'm going to take the[br]C outside the bracket so we

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have the limit as delta x[br]approaches zero of

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C and then our function of x plus delta x[br]takeaway our function of x...

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divided by delta x.

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And because our constant is[br]nothing whatsoever to do with

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our limit is delta x approaches[br]zero we can actually take the

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constant outside of the limiting[br]sign, so the limit delta x approaches 0

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and we have f of x plus delta x, [br]minus our function of x...

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divided by delta x.

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So this is C and this...[br]is our derivative

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so it's actually our f dashed of x.

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So our g dashed of x, our derivative,

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is equal to C times our f dashed of x.

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Now let's have a look at what [br]we're going to do when

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we have two functions that[br]are added together.

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So let's say we have [br]f of x, plus, g of x.

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And we want to differentiate[br]them with respect to x.

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So we want the derivative of[br]that function with respect to x.

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Well, quite simply, what we do[br]is we differentiate each part

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separately and add them together.

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So that's the same as [br]df dx plus dg dx.

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Similarly, what happens if we[br]want two functions that are

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subtracted and we want to[br]differentiate those?

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Well, again, the derivative of those[br]functions subtracted...

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It's very straightforward.

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Because what we've got here.

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Is just the same as we've got[br]here. But with this second

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function multiplied by minus one.

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And there we are using the

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constant multiplier rule. So[br]this is just the same as the

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derivative of d of x of f of x plus

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minus one times the derivative of g of x.

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So that's our df dx plus

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our minus one times our [br]derivative there which is

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plus minus one times dg dx.

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Which is just the same as df dx [br]minus dg dx.

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Let's just return to our table.

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And we can write those in.

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So we have our constant multiplier[br]rule: c times our function of x,

0:08:34.812,0:08:38.848
is c times df dx.

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And if we have f of x plus g of x.

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Our derivative is df dx plus dg dx.

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And if they're subtracted f of x[br]minus g of x.

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Then the derivative[br]is df dx minus dg dx.

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Now let's have a look at[br]an example.

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2 x cubed minus 6 cos x

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And we want to find the[br]derivative with respect to x.

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And that equals the derivative of

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2 x cubed with respect to x, minus

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the derivative of 6 cos x [br]with respect to x.

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Now what we have here is the 2, [br]which is,

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we can use the constant multiplier rule. [br]So that we've got actually:

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Twice the derivative of x cubed[br]with respect to x, minus,

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and again here using the constant[br]multiplier rule, we can

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take the six outside.

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So it's six times the derivative[br]of cos x with respect to x.

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So we have twice, now the[br]derivative of x cubed

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with respect to x,

0:10:14.893,0:10:22.215
is 3 x squared minus, 6 times

0:10:22.215,0:10:27.823
the derivative of cos x, which is[br]minus sine x.

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So we have two threes, 6 x squared,[br]minus times minus is

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positive times 6 sine x.

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Let's look now at extending[br]the table further.

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So let's start with our function[br]f of x again.

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And our derivative.

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df dx or f dash of x.

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And this time let's look at sine[br]of mx where m is

0:11:17.576,0:11:19.400
any constant number

0:11:19.400,0:11:21.900
and the cos of mx.

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Then we'll have a look at e to[br]the power of mx.

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And then log to the[br]base e of mx.

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Let's look then at y equals sine[br]mx. Now here we're going to do a

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substitution and instead of mx,[br]we're going to write

0:11:49.125,0:11:51.422
u is equal to mx.

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So therefore our y is[br]equal to sine u.

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And we're going to differentiate[br]u with respect to x.

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So du by dx is equal to m.

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And then we're going to[br]differentiate y with respect to

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u. So dy by du equals

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the derivative of sine u,[br]which is cos u.

0:12:18.460,0:12:23.941
Now dy by dx, and you'll[br]learn this in the future,

0:12:23.941,0:12:27.023
dy by dx is equal to

0:12:27.023,0:12:33.781
equal to dy by du multiplied by[br]du by dx.

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Now this is called differentiating[br]a function of a function

0:12:38.290,0:12:43.979
or differentiating using the chain rule,[br]and it's the subject of another tutorial.

0:12:44.490,0:12:47.669
So we're not going to go[br]through the details of that now.

0:12:47.669,0:12:51.974
What I want you to do[br]is just use it as a formula.

0:12:53.980,0:12:57.887
Let's now substitute then[br]dy by du is equal to

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cos u multiplied by du by dx,[br]which equals m.

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Now we usually write the[br]constant first, so

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it'll be m times the cos of u.

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But we introduced the u and we[br]want actually dy in terms of dx

0:13:19.640,0:13:24.278
in terms of x, dy by dx.[br]So what we need to do is to

0:13:24.278,0:13:29.334
substitute back and instead of[br]writing u we want to write mx.

0:13:29.334,0:13:37.260
So we have as dy by dx equals m[br]times the cos of mx.

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So in effect, what's happened[br]when we found the derivative is

0:13:42.300,0:13:43.970
we've multiplied by m.

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So let's go back and[br]put that in our table.

0:13:48.590,0:13:54.629
The derivative of sine mx[br]is m cos mx.

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And in a similar way, the[br]derivative of cos mx will be

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minus m sine mx again, just like[br]multiplying by the m.

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Let's look now at e[br]to the power of mx.

0:14:17.770,0:14:21.917
So our y equals x to the [br]power of mx.

0:14:21.917,0:14:26.310
Again, we're going to do a substitution

0:14:26.310,0:14:29.460
and let u equal mx.

0:14:31.130,0:14:34.518
So that our y is equal to e[br]to the power u.

0:14:36.060,0:14:40.322
Again, we're going to[br]differentiate u with respect to x.

0:14:40.322,0:14:46.194
I'm gonna get m and will differentiate[br]y with respect to u.

0:14:47.160,0:14:48.490
And we get e to the u.

0:14:50.370,0:15:00.125
Our dy by dx is equal to[br]dy by du multiplied by du by dx.

0:15:01.370,0:15:06.735
And if we substitute in, dy by du is[br]e to the u.

0:15:06.735,0:15:08.590
du by dx is m.

0:15:09.320,0:15:12.608
Again, let's write our[br]constant terms first, so

0:15:12.608,0:15:17.129
we get m e to the power u.[br]Now we need to

0:15:17.129,0:15:24.214
substitute back for our u so[br]we get m e to the power of mx.

0:15:24.570,0:15:30.810
So again, our derivative we've[br]ended up multiplying by m.

0:15:32.030,0:15:33.724
Let's put that back in our table.

0:15:36.320,0:15:40.730
So e to the power of mx.[br]The derivative is

0:15:40.730,0:15:43.376
m e to the power of mx.

0:15:45.240,0:15:49.376
Let's look now at log to the[br]base e of mx.

0:15:53.770,0:15:58.533
So y equals log to the base e[br]of mx.

0:15:58.533,0:16:04.338
Same process again.[br]Let's let u equal mx.

0:16:05.310,0:16:09.710
So our y equals log to[br]the base e of u.

0:16:11.280,0:16:15.410
Our du by dx equals m.

0:16:16.920,0:16:24.653
dy by du, the differential of[br]log to the base e of u is 1 over u.

0:16:26.540,0:16:30.876
So our dy by dx equals

0:16:30.876,0:16:36.010
dy by du multiplied by[br]du by dx.

0:16:37.350,0:16:43.759
Our dy by du is 1 over u.[br]Our du by dx is m.

0:16:45.760,0:16:52.871
So we have m divided by u,[br]which is mx and very

0:16:52.871,0:16:57.858
conveniently here, our m's cancel out

0:16:57.858,0:17:06.808
and we end up with 1 over x.[br]So dy by dx is equal to 1 over x.

0:17:07.720,0:17:09.722
So let's put that now in our

0:17:09.722,0:17:15.735
table. So log to the base E of[br]MX is our function, so our

0:17:15.735,0:17:20.610
derivative is one over X and of[br]course this is only the case

0:17:20.610,0:17:24.735
where M is greater than zero.[br]Since we can't take the

0:17:24.735,0:17:26.610
logarithm of a negative number.

0:17:28.200,0:17:31.440
Let's just go back to the[br]calculation that we've just

0:17:31.440,0:17:35.004
done, because there is another[br]way that we could have looked

0:17:35.004,0:17:39.540
at. Differentiating Y equals log[br]to the base E of MX and that was

0:17:39.540,0:17:41.484
by using the laws of logarithms.

0:17:42.180,0:17:47.028
What we could have done was[br]said that Y equals log to

0:17:47.028,0:17:52.280
the base E of M plus log to[br]the base E of X.

0:17:53.430,0:17:55.370
And then differentiated them.

0:17:56.410,0:18:03.626
So I DY by DX is equal to[br]what log to the base E of M.

0:18:04.290,0:18:09.846
Is just a constant. So when we[br]differentiate that we get 0

0:18:09.846,0:18:14.400
plus. Log to the base E of X[br]where we know the derivative

0:18:14.400,0:18:15.580
is one over X.

0:18:16.810,0:18:22.165
So we end up with just the same[br]as before divided by DX is equal

0:18:22.165,0:18:23.593
to one over X.

0:18:27.770,0:18:32.438
Not one final one to[br]have a look at is log to

0:18:32.438,0:18:35.161
the base E of AX plus B.

0:18:37.290,0:18:39.558
So let's just have[br]a look at that one.

0:18:48.930,0:18:52.269
So Y equals log to the base.

0:18:52.950,0:18:55.536
A of a X plus B.

0:18:57.440,0:19:00.261
Again, we're going to[br]use the substitution.

0:19:00.261,0:19:02.679
You equals a X plus B.

0:19:04.120,0:19:06.937
So why? Why is it cool to log to

0:19:06.937,0:19:09.310
the base E? Of you.

0:19:12.280,0:19:16.272
We differentiate you[br]with respect to X. Do

0:19:16.272,0:19:18.767
you buy DX equals a?

0:19:20.430,0:19:24.800
At the Y fi do you[br]equals one over you?

0:19:26.490,0:19:33.310
Again, I DYIDX is equal[br]to DY by du multiplied

0:19:33.310,0:19:36.720
by to you by DMX.

0:19:38.070,0:19:40.630
The Wi-Fi do you[br]is one over you.

0:19:42.250,0:19:46.669
Multiplied by do you buy DX,[br]which is a?

0:19:48.030,0:19:55.386
So we have a divided by[br]IU, which is a X plus

0:19:55.386,0:19:55.999
B.

0:19:57.670,0:19:59.760
So do why by DX.

0:20:00.180,0:20:03.789
Sequel to a divided by a[br]X plus B.

0:20:04.810,0:20:07.434
So finally, let's add that[br]to our table.

0:20:10.030,0:20:16.232
The derivative of locked the[br]base E of X Plus B is a divided

0:20:16.232,0:20:18.004
by 8X Plus B.