all right in this video we're going to
talk about finding equations of tangent
lines and finding equations of tangent
lines is something they like to ask a
lot in calculus after you started
talking about derivatives so to find the
equation of the tangent line the only
thing you really need to know you need
to recall point-slope formula so this is
an equation of a line with slope M that
goes through the point y1 and x1 so x1
y1 and to get the slope to plug into the
formula that's what the derivative does
you just simply take the derivative plug
in the x-coordinate that you're
interested in finding the slope at and
that's going to give you your slope and
then you'll have your your line your the
equation of the line so suppose when you
find the equation of the tangent line at
the point 2 4 on the graph of y equals x
squared well already I know this point
so it's going to be Y minus 4 and I've
got to figure out the slope but again
the x-coordinate that I'm doing this at
is at 2 well to figure out the slope of
the tangent line again i can write my
function as f of X if I take the
derivative of that F prime of X the
derivative of x squared is going to be 2
at 2 X so the derivative evaluated at
this point this x coordinate of 2 is
simply going to give me 2 times 2
or a slope of 4 and that's what I'm just
going to simply plug plug right back
into my equation so it says the equation
of the tangent line is going to be y
equals so if I plug 4 in I'll get 4x 4
times negative 2 is negative 8 I'll add
the 4 from the less left side over and
get positive 4 and I'll get the line for
X minus 4 being my tangent line and if
you think about this this seems
reasonable if you think about the graph
of y equals x squared it's just a
parabola so maybe here's the point 2
comma 4 not quite to scale but this
thing could definitely have a tangent
line with a slope of positive 4 okay so
that's really the only tricky part of
these problems is simply just taking the
derivative in the first place so let's
do some more complicated ones of this so
again in all these we're going to find
the equation of the tangent line suppose
we want to find the equation of the
tangent line that lies on the graph of
2x over X plus 1 and we want to do this
at the point 1 comma 1 well again in
this one we know the point it goes
through the only thing we really have to
work to find is the slope of the tangent
line so to figure out the slope of the
tangent line since I have a fraction
I'll use the quotient rule so it says
you write the bottom thing and then you
multiply that by the derivative of the
top you put a minus sign in there says
you leave the top part alone and then
you take the derivative of the bottom
part and this all goes over the bottom
part squared so I'm going to simplify
this down it looks like if you
distribute you'll get a 2x plus 2 on top
minus 2x all over X plus 1 quantity
squared
so my two exes are going to cancel out
and I'll simply have to over X plus one
squared and again in this case to figure
out the slope of the tangent line the
slope of the tangent line comes from the
x coordinate of one so I'll simply plug
one into my derivative and it says I'm
going to get two over well in this case
one plus one squared that's going to
give me two over two squared which is
going to give me one-half so it says the
equation of my tangent line again is
going to be y minus the y coordinate
which was one and that's going to equal
the slope of the tangent line which we
found from taking the derivative and
then X minus the x coordinate that was
given and there is the equation of my
tangent line and unless it really
specifies and says put it in slope
intercept form I'll just leave it like
this unless there's a good reason not to
I mean there's the equation of a tangent
line
let's maybe do at least one or two more
here suppose I have the function y
equals one plus two x quantity squared
and we want to find the equation of the
tangent line at the point 1 comma 9
alright well in this case again I'm
going to have to take my derivative
again we could call Y that's the same
thing as f of X so I'll rewrite f prime
of X and in this case we're going to
have to use the chain rule so the two
will come out front
we'll leave the inside stuff alone
we'll take one away from the power and
then we have to multiply by the
derivative of the inside stuff the
derivative of the inside the derivative
of 1 is zero the derivative of positive
2x is just positive two we'll multiply 2
and 2 and get 4 and then have 1 plus 2x
left over so again we're interested in
finding the slope of the tangent line at
the x coordinate of 1 so we'll plug one
end and it says you'll get 4 times 1
plus well 2 times 1
so that's four times it looks like the
inside you get one plus two which is
three and all of that simply is going to
work out to twelve so again the equation
of your tangent line you'll have y minus
the y-coordinate equals the slope X
minus the x-coordinate again the
y-coordinate we were given is this value
of nine we found the slope of the
tangent line being twelve the
x-coordinate was one so there's the
equation of your tangent line maybe
let's do one last one here suppose we
want to find the equation of the tangent
line to the graph y equals tangent of x
at the x coordinate of PI over four so
here they don't give us the y coordinate
but it's easy enough well we can figure
out the y coordinate by simply plugging
PI over four into the graph so f of PI
over four that's going to be tangent of
PI over four and tangent of PI over four
turns out to be one so the point that
we're using is going to be PI over four
comma one and again now to find the
slope of the tangent line I'll simply
take the derivative of my function the
derivative of tangent X is secant
squared of X so I'm going to have to
evaluate secant squared at this value PI
over four so PI over four
and let's see so so I've got secant
squared of PI over 4 and remember we can
rewrite secant squared as 1 over cosine
squared of PI over 4 well cosine of PI
over 4 is the value square root of 2
over 2
but again that's all being squared here
so that's going to give us 1 over 2 over
4 so if I flip and multiply
I'll get 4 over 2 or 2 as the slope of
my tangent line so to put it all
together again it says I'll take Y minus
the y coordinate that will equal the
slope of the tangent line times X minus
the x-coordinate
well again the y coordinate we found was
positive 1 the slope of the tangent line
we found was positive 2 and the x
coordinate we said is this value PI over
4 and again there would be the equation
of your tangent line in this case so
some basic examples really all it comes
down to is simply taking derivatives if
you also need to see finding equations
of tangent lines if your function is
defined implicitly that's when you have
X's and Y's all mixed up feel free to
take a look at my website I have some
problems where you do implicit
differentiation and you also find
equations of tangent lines for an
implicitly defined function if you have
any questions shoot me an e-mail I'll be
happy to get to them