0:00:00.860,0:00:02.500
So we have the[br]function f of x is

0:00:02.500,0:00:05.290
equal to e to the[br]x over x squared.

0:00:05.290,0:00:07.630
And what I want to[br]do in this video

0:00:07.630,0:00:10.960
is find the equation,[br]not of the tangent line,

0:00:10.960,0:00:17.220
but the equation of the normal[br]line, when x is equal to 1.

0:00:17.220,0:00:22.970
So we care about the[br]equation of the normal line.

0:00:22.970,0:00:26.250
So I encourage you to pause this[br]video and try this on your own.

0:00:26.250,0:00:29.690
And if you need a little bit[br]of a hint, the hint I will give

0:00:29.690,0:00:33.800
you is, is that the[br]slope of a normal line

0:00:33.800,0:00:38.920
is going to be the[br]negative reciprocal

0:00:38.920,0:00:41.020
of the slope of[br]the tangent line.

0:00:41.020,0:00:46.640
If you imagine a[br]curve like this,

0:00:46.640,0:00:49.686
and we want to find a[br]tangent line at a point,

0:00:49.686,0:00:51.310
it's going to look[br]something like this.

0:00:51.310,0:00:54.600
So the tangent line is[br]going to look like this.

0:00:54.600,0:00:59.220
A normal line is perpendicular[br]to the tangent line.

0:00:59.220,0:01:01.470
This is the tangent line.

0:01:01.470,0:01:05.710
The normal line is going to[br]be perpendicular to that.

0:01:05.710,0:01:08.310
It's going to go just like that.

0:01:08.310,0:01:11.450
And if this has a[br]slope of m, then this

0:01:11.450,0:01:14.100
has a slope of the[br]negative reciprocal of m.

0:01:14.100,0:01:17.350
So negative 1/m.

0:01:17.350,0:01:19.200
So with that as a[br]little bit of a hint,

0:01:19.200,0:01:21.880
I encourage you to find the[br]equation of the normal line

0:01:21.880,0:01:25.410
to this curve, when x equals 1.

0:01:25.410,0:01:29.082
So let's find the slope[br]of the tangent line.

0:01:29.082,0:01:30.790
And then we take the[br]negative reciprocal,

0:01:30.790,0:01:33.070
we can find the slope[br]of the normal line.

0:01:33.070,0:01:35.190
So to find the slope[br]of the tangent line,

0:01:35.190,0:01:38.660
we just take the derivative here[br]and evaluate it at x equals 1.

0:01:38.660,0:01:41.180
So f prime of x,[br]and actually, let me

0:01:41.180,0:01:42.400
rewrite this a little bit.

0:01:42.400,0:01:47.222
So f of x is equal to e to the[br]x times x to the negative 2.

0:01:47.222,0:01:49.180
I like to rewrite it this[br]way, because I always

0:01:49.180,0:01:50.721
forget the whole[br]quotient rule thing.

0:01:50.721,0:01:52.770
I like the power[br]rule a lot more.

0:01:52.770,0:01:54.610
And this allows me to[br]use the power rule.

0:01:54.610,0:01:57.792
I'm sorry, not the power[br]rule, the product rule.

0:01:57.792,0:01:59.250
So this allows me[br]to do the product

0:01:59.250,0:02:01.150
rule instead of[br]the quotient rule.

0:02:01.150,0:02:03.650
So the derivative[br]of this, f prime

0:02:03.650,0:02:05.930
of x, is going to be the[br]derivative of e to the x.

0:02:05.930,0:02:11.500
Which is just e to the x times[br]x to the negative 2, plus

0:02:11.500,0:02:14.780
e to the x times the derivative[br]of x to the negative 2.

0:02:14.780,0:02:18.650
Which is negative 2x to[br]the negative 3 power.

0:02:18.650,0:02:21.200
I just used the power[br]rule right over here.

0:02:21.200,0:02:24.710
So if I want to evaluate[br]when x is equal to 1,

0:02:24.710,0:02:26.450
this is going to[br]be equal to-- let

0:02:26.450,0:02:28.570
me do that in that[br]yellow color like.

0:02:28.570,0:02:30.000
I like switching colors.

0:02:30.000,0:02:32.950
This is going to be[br]equal to, let's see,

0:02:32.950,0:02:35.580
this is going to be[br]e to the first power.

0:02:35.580,0:02:39.100
Which is just e times[br]1 to the negative 2,

0:02:39.100,0:02:46.290
which is just 1 plus e to the[br]first power, which is just e,

0:02:46.290,0:02:48.410
times negative 2.

0:02:48.410,0:02:50.210
1 to the negative 3 is just 1.

0:02:50.210,0:02:51.644
So e times negative 2.

0:02:51.644,0:02:52.810
So let me write it this way.

0:02:52.810,0:02:57.390
So minus 2e.

0:02:57.390,0:03:02.220
And e minus 2e is just going[br]to be equal to negative e.

0:03:02.220,0:03:12.190
So this right over here, this is[br]the slope of the tangent line.

0:03:12.190,0:03:14.390
And so if we want the[br]slope of the normal,

0:03:14.390,0:03:16.100
we just take the[br]negative reciprocal.

0:03:16.100,0:03:19.702
So the negative reciprocal[br]of this is going to be,

0:03:19.702,0:03:21.410
well the reciprocal[br]is 1 over negative e,

0:03:21.410,0:03:22.785
but we want the[br]negative of that.

0:03:22.785,0:03:24.240
So it's going to be 1/e.

0:03:24.240,0:03:27.260
This is going to be the[br]slope of the normal line.

0:03:34.590,0:03:37.066
And then if we, and[br]our goal isn't just

0:03:37.066,0:03:38.524
to the slope of[br]the normal line, we

0:03:38.524,0:03:40.340
want the equation[br]of the normal line.

0:03:40.340,0:03:42.560
And we know the[br]equation of a line

0:03:42.560,0:03:45.860
can be represented[br]as y is equal to mx

0:03:45.860,0:03:47.970
plus b, where m is the slope.

0:03:47.970,0:03:52.050
So we can say it's going to be[br]y is equal to 1/e-- remember,

0:03:52.050,0:03:56.660
we're doing the normal[br]line here-- times x plus b.

0:03:56.660,0:03:59.250
And to solve for b, we[br]just have to recognize

0:03:59.250,0:04:01.210
that we know a point[br]that this goes through.

0:04:01.210,0:04:04.180
This goes through[br]the point x equals 1.

0:04:04.180,0:04:09.970
And when x equals 1, what is y?

0:04:09.970,0:04:13.430
Well, y is e to the 1st[br]over 1, which is just e.

0:04:13.430,0:04:16.820
So this goes to the[br]point 1 comma e.

0:04:16.820,0:04:22.800
So we know that when x is[br]equal to 1, y is equal to e.

0:04:22.800,0:04:25.020
And now we can just solve for b.

0:04:25.020,0:04:34.960
So we get e is[br]equal to 1/e plus b.

0:04:34.960,0:04:38.630
Or we could just subtract[br]1 over e from both sides,

0:04:38.630,0:04:45.160
and we would get b is[br]equal to e minus 1/e.

0:04:48.190,0:04:53.355
And we could obviously right[br]this as e squared minus 1/e

0:04:53.355,0:04:54.730
if we want to[br]write it like that.

0:04:54.730,0:04:56.355
But could just leave[br]it just like this.

0:04:56.355,0:04:58.170
So the equation of[br]the normal line--

0:04:58.170,0:05:00.630
so we deserve our drum[br]roll right over here--

0:05:00.630,0:05:10.020
is going to be y is equal[br]to 1/e times x, plus b.

0:05:10.020,0:05:13.400
And b, plus b, is all of this.

0:05:13.400,0:05:18.830
So plus e minus 1/e.

0:05:18.830,0:05:25.530
So that right there is our[br]equation of the normal line.