Trigonometric Derivatives - Overview


Uploaded by TheIntegralCALC on 19.07.2011

Transcript:
Hi, everyone! Welcome back to integralcalc.com. Today we’re going to be talking about trigonometric
derivatives. And I just wanted to give you an overview of how you’re going to work
with trigonometric derivatives. We’re going to talk about four things primarily. The first
is your basic sine and cosine derivatives and the next thing will be the other trigonometric
identities: tangent, secant, co-tangent, and co-secant. And finally, we’ll talk about
how you oft4en use trigonometric derivatives or you apply chain rule to trigonometric derivatives
and how to find higher-order trigonometric derivatives.
Let’s talk about the first thing here. I’ve got sine of x here and what I wanted to point
out to you guys was that if you don’t know anything else about trigonometric derivatives,
the first thing that you want to know  is that the derivative of sine of x is cosine
of x. And then that the derivative of cosine of x is negative sine of x. I’m hoping to
see that we’re going to start to see a pattern here. The derivative of negative sine of x
is negative cosine of x and the derivative of negative cosine of x brings you back to
sine of x. So what I wanted to point out was that if you start with sine of x and you can
memorize this sequence that the derivative of sine of x is cosine of x then that you
flip to negatives and you get negative sine and a negative cosine. And then that the derivative
of negative cosine is back to sine and this pattern repeats itself. If you memorize that
sequence of derivatives, it’s going to help you immensely when you’re trying to take
the derivatives of the other trigonometric identities and apply this to real problems.
So if anything about trigonometric derivatives, memorize this sequence.
Now we’re going to talk about the other trigonometric identities: tangent, secant,
cotangent and cosecant. The reason it’s so important to memorize these derivatives
is because you can break tangent, secant, co-tangent and co-secant into values that
are defined just by sine and cosine. So obviously, tangent of x is the same thing as sine of
x over cosine of x. Secant of x is 1 over cosine of x. Co-tangent of x is the same thing
as cosine of x over sine of x and cosecant of x is 1 over sine of x. So all of these
other trigonometric identities are defined by sine and cosine in some way. So if you
know that, you can always derive these derivative formulas. So first, memorize these guys over
here. Second, if you can memorize these over here but the derivative of tangent of x is
secant squared of x. That the derivative of secant of x is secant of x times tangent of
x. That the derivative of co-tangent of x is negative cosecant squared of x. memorize
this set if you can. But if you can’t memorize it, you can always derive all of these derivatives
using your knowledge of these first set and quotient rule. So if you know quotient rule
and you’re comfortable with that, you can figure out any of these. Notice that everything
here is a quotient so you can take tangent, secant, co-tangent and co-secant and change
them into a quotient, a fraction and then you can use quotient rule to find the derivative.
You can do it with any of these but we’re just going to do co-secant of x to show you.
Remember, if you’re taking the derivative of a fraction and you’re going to use quotient
rul to do it, quotient rule tells you that you need first the derivative of the numerator.
The derivative of 1 is just zero because the derivative of any constant is zero. Then you
multiply by the denominator so you’re going to multi9ply by sin of x, then you subtract
and you put the numerator then you multiply by the derivative of the denominator. The
derivative of sin of x, if we go and look back over here, we see that the derivative
of sine of x is co-sine of x so you multiply by cosine of x. And then according to quotient
rule, you divide by the square of the denominator so you’ve got sine of x squared in our denominator.
And now, if we simplify this, obviously in the numerator here, zero times sine of x is
just going to be zero. This whole thing is going to cancel so we’re going to be left
with negative cosine of x in the numerator and sine of x squared in the denominator.
We can just change that into the following: If we split apart the denominator, sine of
x squared into sine of x times sine of x, we can split apart this fraction like this.
We can get negative cosine of x divided by sine of x and then we multiply by 1 over sine
of x. Notice how that gives us negative cosine of x in the numerator and sin squared of x
in the denominator. The reason we split apart this fraction is because notice this first
fraction. We have cosine of x over sine of x. We see that that’s right here. Cosine
of x over sine of x. Because we’ve got this negative sign right here, we know that this
negative cosine of x over sine of x is negative co-tangent of x. And then 1/(sin x) here,
we already know to be co-secant of x. We’ve got (–cot x)(cosec x). And this is in fact
the derivative of cosecant of x. Notice that you can use quotient rule and your knowledge
of these first set of derivatives to find any of the derivatives of these other trigonometric
identities. So now that we know what the derivatives of
all of our trigonometric identities are, let’s talk about how they’re often used in regular
problems. So first, let’s talk about chain rule. We’ve got this function here: y = 2sin
10x + 3cos pi x. We use chain rule a lot with trigonometric derivatives because inside the
sine function here, we’ve got 10x. Not just regular x like we’ve got over here on this
sine of x function but 10x. Whenever you have something other than the normal variable x
inside a trigonometric identity, you’re going to use chain rule to take its derivative.
So let’s take the derivative of this term by term. The derivative of 2sin 10x, we’re
going to leave the 2 as it is and then the derivative of sine is cosine. So we will take
the derivative cosine but according to chain rule, we’ll leave the 10x right where it
is but we have to multiply this whole thing by the derivative of the inside here. So we’re
going to multiply by the derivative of 10x. We multiply it by 10. Same thing here with
3cos pi x. We get 3. We know the derivative of cosine to be negative sine. So we go ahead
and multiply that by negative sine of pi x. but then because we’ve got to multiply by
the derivative of what’s inside of the cosine function, the derivative of pi x is just pi.
We have to multiply it by pi. So when we simplify this, we can se that we get 20 cos 10x – 3pi
sin pi x. So that’s how we take the derivative there. Remember with chain rule, you take
the derivative of the outside first so we take the derivative of the entire sine function,
leaving the inside function 10x completely alone. But then we had to multiply by the
derivative of the inside so we multiplied by 10 and we did the same thing with the 3cos
pi x. So that’s how we use chain rule in conjunction with trigonometric derivatives.
Now, let’s talk about what we do when we’ve got higher-order trigonometric derivatives.
We’ve got in this problem (sin^2 3x)(cos^4 5x). So what I always recommend doing with
these higher-order derivatives is rewriting them because I think it’s a lot easier to
look at. I don’t get confused as easily if I move the exponents. So sin^2 3x is the
same thing as saying (sin 3x)^2 times (cos 5x)^4. Again, you’ll need chain rule. We’ve
got kind of three nested functions here. We have the full function, then we’ve got an inner
function sin 3x and then we’ve got a third inner function which is just this 3x. So according
to chain rule, we have to take the derivative of the outside function, leaving all of the
inner functions alone and hen work our way in. We’ll also need to use product rule
for this one because we’ve got one function here (sin 3x)^2 and then the other function
(cos 5x)^4. So according to product rule, we’re going to take the derivative of (sin
3x)^2 and then we’re going to multiply that by (cos 5x)^4 and then we’ll add to that
the derivative of (cos 5x)^4 and multiply it by (sin 3x)^2. So now, let’s go ahead
and start taking our derivatives. This was just our set-up for product rule but let’s
start taking our derivatives. So looking at our functions here, we will
just take the derivative of the outside. That requires us to bring the 2 out in front so
we get 2 (sin 3x), because we’re going to leave sin 3x completely alone and just take
the derivative of the outside. So we bring the exponent out in front and then subtract
1 from the exponent. We don’t have to write that exponent 1 because it’s just redundant.
But now according to chain rule, we’ll have to multiply by the derivative of the inside
functions. So the derivative of sin 3x is just going to be cos 3x. We know that the
derivative of sine is cosine so we multiply by cos 3x but then according to chain rule,
we still have to multiply by the derivative of this inner function 3x. So, the derivative
of 3x is just 3. We go ahead and multiply by 3. That’s the derivative of that one.
Of course, we still have to multiply by (cos 5x)^4. An now we can go ahead and take the
derivative of (cos 5x)^4. Using chain rule again, we bring the 4 out in front we’ll
get 4(cos 5x)^3 then we multiply by the derivative of cos 5x which is going to be –sin 5x.
But then again, we have to multiply it by the derivative of 5x. So we multiply it by
5. And then of course multiplied by (sin 3x)^2. So when we simplify that, for our first term
ehre, we’ll get 6(sin 3x)(cos 3x)(cos^4 5x). We’ve got (cos 5x)^4. I would recommend
moving this exponent back in here between cos and 5x because that’s technically the
most correct way to write it. So cos^4 5x. And then for our second term, we have minus
20(cos^3 5x)(sin 5x)(sin^2 3x). So again, with higher-order trigonometric
derivatives, move the exponents to the outside and then use chain rule to work your way from
the outside in to get those derivatives then move the exponents back in when you’re done
with everything. So anyway, I hope that video helped you guys. I hope it gave you an overview
of how to work with trigonometric derivatives and some of the common things you run into.
I hope it helped and I will see you in the next video. Bye!