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!

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!