PROFESSOR: Welcome back to recitation.
In this segment I'm going to actually show, well, you're
actually going to show, the derivative of tangent x using
the quotient rule.
So what I'd like you to do, I wanted to remind you of what
the quotient rule is.
So u and v are functions of x.
We want to take the derivative of u divided by v. I've
written the formula that you were given in class for this.
And I'm asking for you to take d dx of tangent x using the
quotient rule.
And the hint I will give you is the reason we can obviously
use the quotient rule is because tangent x is equal to
a quotient of two functions of x.
It's sine x divided by cosine of x.
So I'm going to give you a minute to work this out for
yourself, and then when we come back I
will do it for you.
OK.
So we want to find the derivative of tangent x.
So let me, let me work on this side of the board.
So I'm actually going to take d dx of sine x divided by
cosine of x.
OK.
So in this case, sign x is u.
Cosine x is v. So using my quotient rule I know that
first I have to take the derivative of sine x--
that's cosine x-- and then I multiply it by the
denominator, the v, which is cosine x.
So my first term in the numerator is cosine squared x.
Again, one cosine x comes from the derivative of sine x, one
cosine x is the v. It's the cosine x in the denominator.
Then I have to subtract v prime u.
The derivative of cosine x is negative sine x.
I'll actually just write that one down.
And then I bring the u along for the ride.
So I multiply by sine x here.
And then I take v squared in the
denominator from the formala.
v, again, is cosine x, so I take cosine squared x in the
denominator.
Now this at this point is a little bit messy, but the nice
thing is that we can use some trigonometric identities to
simplify this.
So let me first write out what it is a little more clearly.
Minus a negative gives you a positive.
And then here I get sine x times sine x, so I get the
sine squared x.
And then I keep divided by cosine squared x.
Now at this point some of you might have divided by cosine
squared x here and gotten 1, and divided by cosine squared
x here and gotten tangent squared x.
And then from there you could simplify to another
trigonometric function.
I'm going to go straight a different way to show you what
that actually also equals.
So there, at this point I want to stress there are sort of
two ways you can get to the same place.
But I'm going to use the fact that the numerator is a very
nice trigonometric identity that we know.
We know cosine squared x plus sine squared x
always equals 1.
So this is quite lovely, the numerator simplifies to 1, the
denominator stays cosine squared x.
What is this function?
1 over cosine x is actually secant x.
So if you need at this point to rewrite the whole thing
like this, right?--
1 squared is 1 and in the denominator we still get
cosine squared x--
this tells you that 1 over cosine squared x is actually
just equal to secant squared x.
So again, what I want to point out is we've now taken the
derivative of tangent x and we got that that's
secant squared x.
Now using this quotient rule, you can do the same kind of
thing with cotangent x, with cosecant x, with secant x.
You can find all these derivatives of these
trigonometric functions using the quotient rule.
So if you want to know what the derivative of secant x is,
you should take dd x of 1 divided by cosine x and use
the quotient rule.
Or, in fact, the chain rule would work well there also, to
find that derivative.
So we are building up the number of derivatives we can
find using these different rules.
So we'll stop there.