Equation of a tangent line

Finding the equation of the line tangent to f(x)=xe^x when x=1

I've told you multiple times
that the derivative of a curve

at a point is the slope of the
tangent line, but our

friend [? Akosh ?]

sent me a problem where it
actually wants you to find the

equation of the tangent line.

And I realize, I've never
actually done that.

So it's worthwhile.

So let's do that.

So it says, find the equation
of the tangent line to the

function f of x is equal to x e
to the x at x is equal to 1.

So let's just get an intuition
of what we're even looking for.

So this function is going to
look something like, I actually

graphed it, because it's not
a trivial function to graph.

So this is x e to the x,
this is what it looks like.

I'm just using a graphing
calculator, and you can

see, I just typed it in.

And what this is
asking us, is ok.

At the point, x is equal to 1.

So this is the point
x is equal to one.

So f of x is going to be
someplace up here, and

actually, f of x is going
to be equal to e, right?

Because f of 1 is
equal to what?

1 times e to the 1.

So it equals e.

So we're saying at the point,
at the point 1 comma e, so at

the point 1 comma 2.71,
whatever, whatever.

So that's what point?

That's this point.

So it's right here.

2 point, this is e right
here, the point 1 comma e.

So we want to do is figure
out the equation of the

line tangent to this point.

So what we're going to do, is
we're going to solve it by

figuring out its slope, which
is just the derivative

at that point.

So we have to figure
out the derivative at

exactly this point.

And then we use what we learned
from algebra 1 to figure out

its equation, and we'll graph
it here, just to confirm that

we actually figured out the
equation of the tangent line.

So the first thing we want to
know is the slope of the

tangent line, and that's just
the derivative at this point.

When x is equal to 1, or
at the point 1 comma e.

So what's the
derivative of this?

So f prime of x.

f prime of x is equal to,
well, this looks like a

job for the product rule.

Because we know how to figure
out the derivative of x, we

know how to figure out the
derivative of e to the x, and

they're just multiplying
by each other.

So the product rules help us.

The derivative of this thing
is going to be equal to the

derivative of the first
expression of the

first function.

So the derivative of x is just
1, times the second function,

times e to the x, plus the
first function, x, times the

derivative of the
second function.

So what's the derivative
of e to the x?

And that's what I find so
amazing about the number e, or

the function e to the x, is
that the derivative of e

to the x is e to the x.

The slope at any point of
this curve is equal to the

value of the function.

So this is the derivative.

So what is the derivative of
this function at the point x

is equal to 1, or at
the point 1 comma e?

So we just evaluate it.

We say f prime of 1 is equal to
1 time e to the 1 plus 1 times

e to the 1, well, that's
just equal e plus e.

And that's just equal to 2 e.

And you know, we could figure
out what that number, e is just

a constant number, but we write
e because it's easier to write

e than 2.7 et cetera, and an
infinite number of digits,

so we just write 2e.

So this is the slope of the
equation, or this is the slope

of the curve when x is equal to
one, or at the point

1e, or 1 f of 1.

So what is the equation
of the tangent line?

So let's go ahead and take this
form, the equation's going to

be y is equal to, I'm just
writing it in the, you know,

not the point slope, the mx
plus b form that you

learned in algebra.

So the slope is going to be 2e.

We just learned that here.

That's the derivative
when x is equal to 1.

So 2e times x plus
the y-intercept.

So if we can figure out
the y-intercept of this

line, we are done.

We have figured out the
equation of the tangent line.

So how do we do that?

Well, if we knew a y or
an x where this equation

goes through, we could
then solve for b.

And we know a y and x that
satisfies this equation.

The point 1 comma e.

The point where we're trying to
find the tangent line, right?

So this point, 1 comma e,
this is where we want to

find the tangent line.

And by definition, the
tangent line is going to

go through that point.

So let's substitute those
points back in here, or this

point back into this equation,
and then solve for b.

So y is equal to e, is equal to
2 e, that's just the slope at

that point, times x,
times 1, plus b.

It might confuse you, because
e, you'll say, oh, e,

is that a variable?

No, it's a number,
remember, it's like pi.

It's a number.

You can substitute 2.7 whatever
there, but we're not doing

that, because this is cleaner.

And let's solve.

So you get e is
equal to 2e plus b.

Let's subtract 2e
from both sides.

You get b is equal
to e minus 2e.

b is equal to minus e.

Now we're done.

What's the equation
of the tangent line?

It is y is equal to
2 times e x plus b.

But b is minus e,
so it's minus e.

So this is the equation
of the tangent line.

If you don't like these e's
there, you could replace that

with the number 2.7 et cetera,
and this would become 5 point

something, and this would
just be minus 2.7 something.

But this looks neater.

And let's confirm.

Let's use this little graphing
calculator to confirm that that

really is the equation
of the tangent line.

So let me type it in here.

So it's 2, 2 times e times x,
right, that's 2ex minus e.

And let us graph this line.

There we go.

It graphed it.

And notice that that line, that
green line, I don't know if you

can, maybe I need to make this
bigger for it to

show up, bolder.

I don't know if that helps.

But if you look here, so this
red, this is our original

equation, x e to the
x, that's this curve.

We want to know equation
of the tangent line

at x is equal to 1.

So it's the point
x is equal to 1.

And when x is equal to 1, f of
x is e, right, you can just

substitute back into the
original equation to get that.

So this is the
point, 1 comma e.

So the equation of the tangent
line, its slope is going to be

the derivative at this point.

So we solved the derivative of
this function, and evaluated

it at x is equal to 1.

That's what we did here.

We figured out the derivative,
evaluated x equals 1.

And so we said, OK, the slope.

The slope at when x is equal to
1 and y is equal to e, the

slope at that point
is equal to 2e.

And we figured that out
from the derivative.

And then we just used our
algebra 1 skills to figure out

the equation of that line.

And how did we do that?

We knew the slope, because
that's just the derivative

at that point.

And then we just have to
solve for the y-intercept.

And the way we did that is we
said, well, the point 1 comma e

is on this green line as well.

So we substituted that in, and
solve for our y-intercept,

which we got as minus e, and
notice that this line

intersects the y-axis at minus
e, that's about minus

2.7 something.

And there we have it.

We have shown that, and
visually, it shows that

this is the tangent line.

Anyway, hope you found
that vaguely useful.

If you did, you should
thank [? Akosh ?]

for being unusually persistent,
and having me do this problem.

See you in the next video.