Bam!
Mr. Tarrou.
In this video, I'm wanna talk about the six
rules for finding derivatives of trig functions
and memorizing them.
Through your study of calculus, I'm sure you're
going to be, like my students, given a long
list of rules for finding derivatives—and,
ultimately, we'll be talking about integration
fairly soon—that you'll have to memorize
to be successful on your test and certainly
if you're gonna take the AP exam, as well.
And I'm, you know, quizzing my students right
now on their memorization of the six trig
rules for finding derivatives.
And they're doing fine, but I'm a little bit
worried that at the end of the year, when
they have a whole list of rules to have memorized...
You know, what happens when the nerves get
to you on the test?
And you think, “Okay.
Well, I know the derivative of sine is cosine,
and the derivative of cosine is negative sine
of x,” assuming we're not, you know, needing
the chain rule.
And you get to a problem when you have to
find, let's say, the derivative of tangent
with respect to x.
So the derivative of tangent of x with respect
to x.
Well, right now we're just talking and doing
a lot of work with, again, these derivatives
with these trig functions, and we know right
away that that is equal to secant squared
x.
But, okay, it's a little bit further down
the road in the class—maybe a couple months
down the road.
We're doing a lot of other things, and we
get to a problem, and we're just not quite
sure.
“Was that secant squared, or was it cosecant
sqaured or something else?”
Well, kind of like when we did out trig proofs
in pre-calculus, sometimes we just need to
go back to the parent functions, you know,
our sine and cosine functions—basically
build the other four trig functions.
So I can rewrite this as the derivative with
respect to x of sine of x over cosine of x
and then, you know, walk through, or go through,
the quotient rule of finding this derivative.
And that's going to be the denominator times
the derivative of the numerator (the derivative
of sine is cosine) minus the numerator times
the derivative of the denominator, which is
gonna be derivative of with respect to x of
cosine of x is negative sine x, all over the
denominator squared, which you can write as
cosine x squared.
Or you can write cosine squared of x.
Either way you write that, it's equivalent
statements.
Though I am finding that my students are working
through the chain rule more accurately when
they write cosine squared of x with that power
of 2 on the outside.
At any rate, cosine times cosine is cosine
squared.
Negative times negative is positive.
And that's over, again, that denominator of
cosine squared x.
Now, sine squared plus cosine squared—that's
one of our three Pythagorean identities for
our trig functions, and that's equal to 1.
And one over cosine squared is secant squared.
And, okay, yes...
Obviously a lot quicker if you just remember
that the derivative with respect to x of the
tangent of x is secant squared x, but, you
know, sometimes our nerves get to us on our
test.
There's plenty of other things we have to
have memorized.
If that slipped your mind, you can, you know,
again, write your tangent function in terms
of sine and cosine, go through the quotient
rule, and double check that your memory serves
you right.
And the derivative, again, of tangent is secant
squared x.
And, of course, we can do that same kind of
work through the cotangent function.
But I'm not gonna show you that again.
Right?
It's just gonna be the quotient rule again.
How about if you need to find the derivative
with respect to x of... whoops, not equals
to but of secant of x.
I don't know why I'm wrapping this in parenthesis.
I don't really need that there.
But, actually, I do often like to wrap the
angle that the trig function is being applied
to in parenthesis.
So, either way, the derivative with respect
to x of secant of x.
Well, do we remember what that is?
If not, we can rewrite this as 1 over cosine.
So the derivative with respect to x of 1 over
cosine of x; and we can use the quotient rule
again to find the derivative of this, or we
can write it with a... bringing the cosine
out of the denominator and writing it with
an exponent of –1.
So the derivative with respect to x of cosine
of x to the –1.
Again, I'm bringing that cosine up out of
the denominator, and I'm just gonna wrap the
entire cosine function in parenthesis and
put the –1 out front because I see my students,
you know, finding that, again, they use the
chain rule more correctly when they wrap the
whole trig function in a set of parenthesis.
And I can write this as, you know, cosine
to the –1 of x, but, again, learning from
how my students are seeing this.
And kind of looks like maybe...
You can even wrap that in a set of parenthesis,
as well, if it's bothering you that we're
finding the derivative with respect to x of
all of this, and then the cosine has that
exponent of –1.
And now, I'm just going to use the general
power rule, or chain rule, to work this problem
out.
So we have—bring the exponent down—negative
cosine of x.
And, of course, when I'm doing the general
power rule, I bring this exponent down (and
I can write just a negative sign or you can
write –1 here), reduce the exponent by one,
and now, we're going to multiply by the derivative,
again, with respect to x of the inside function.
See, the cosine of x is the inside function
of this power of –1.
So, you know, times the derivative of cosine
of x—that inside function.
Okay.
Well, now this is gonna become a couple of
things.
I'm gonna go ahead and...
I'm gonna take this cosine of x to the –2,
or this cosine to the –2 power x, and bring
it to the denominator.
And we're just gonna just go ahead and leave
this negative up here for a second.
That might very well cancel out.
Maybe.
The derivative of cosine is negative sine,
and, indeed, we have a negative coming in.
Again, the derivative of cosine is negative
sine x.
And we have a negative out here in the beginning
of the problem, and those are gonna cancel
out, and now we have sine over cosine squared.
Well, that doesn't seem to... remember the
derivative with respect to x of secant of
x is, so let's write it like this.
Ah, okay, so we have...
Sine over cosine is equal to tangent, and,
if I show this multiplication of 1 up here,
we have 1 over cosine is equal to secant.
So, if you have forgotten the derivative with
respect to x of secant of x is equal to secant
x tangent x or tangent x times secant x, however
it's written in your textbook.
And, again, this is longer than it would have
taken if you had just memorized that, but,
if you forget, if you're not quite sure, go
into terms of sine, or cosine in this particular
problem, walk through the derivative process.
In this case, again, I wrote that with a negative
exponent, went through the general power rule,
the chain rule, and got our final answer and
derived the derivative rule on my own, and
you can do this with the other two trig functions.
I'm gonna go ahead and step off now, show
you those two other derivative rules just
for the heck of it, and we'll call this video
done.
And here are your six rules for finding the
derivatives of trig functions.
By the way, if you're gonna go through the
trouble of memorizing them and, you know,
if need be, deriving your own rules, make
sure you copy your answer down correctly.
I did eventually catch the fact that I forgot
that was secant squared.
Just wrote secant.
So I'm Mr. Tarrou.
Bam!
Go do your homework.
