Hey guys!
Welcome to this video on derivatives of Trigonometric Functions.
Now, you don’t take the derivative of a trig function any differently than you would any other function.
Trig functions are just scarier.
So, we thought we’d make a video.
Now, while you still use the same rules to take derivatives of trig functions as you
would for any other function, there ARE a few facts to keep in mind, and derivatives
that you should memorize when it comes to trig functions.
Limθ→0 (sin𝛳/𝛳) = 1 Limθ→0 ((cos𝛳-1)/𝛳) = 0
Being able to spot these in a function can help to make your differentiating much simpler.
Before we jump into differentiation let’s take a look at a few limit problems and use
our facts to help us solve them.
Lim𝛉→0 (sin𝜭/24𝜭)
So we know that one of our limit rules allows us to move this constant out here in front and then evaluate.
So, then we have.
1/24Lim𝛉→0 (sin𝜭/𝜭)
So now at this point, we should be able to recognize this right here. It's one of our facts.
Lim𝛉→0 (sin𝜭/𝜭) is equal to 1, so that leaves us with
1/24 (1).
Which is equal 1/24.
Let’s look a more complicated one:
lima→0 (sin(4a)/sin(7a))
Well.. This doesn’t look anything at all like either of our facts..
However, we can manipulate it in order to make it do what we need it to.
Sometimes you just have to get creative to make numbers do what you want them to.
Alright, let’s separate this fraction into two separate fractions and take the limit.
Which we know we can do because of our limit rules
=lima→0 ((sin(4a)/1)*(1/sin(7a))
Now, once we get to here, we have to think
of ways to make this look like the fact that we have so that we can simplify it.
Well, we can multiply the whole thing by a/a, since a/a is really just 1 and doesn’t change anything.
Now the reason I'm doing that is if you remember, one of our facts is the
=lima→0 ((sin(4a)/1)*(1/sin(7a))a/a
=lima→0 ((sin(4a)/a)*(a/sin(7a))
What we need to happen now is to have the same coefficient in front of our a’s here.
7a=7a =a
To make that happen we can do pretty much what we just did with the a/a.
Except we will multiply sin(4a)/a by 4/4 and a/sin(7a) by 7/7.
=lima→0 ((4sin(4a)/4a)*(7a/7sin(7a))
Then, we can move these constants out in front of our limit,
and if you recall one of our limit properties tells us that we can break up this property into separate limits,
find the limits, and then put them back together.
= 4lima→0 ((sin(4a)/4a) 1/7lima→0(7a/sin(7a))
Now, that the coefficients in front of our a’s here
are the same we know that these both are equal to 1.
So, now we just have
=4(1/7) which is equal to 4/7
Let’s look at one more before we move onto differentiation.
Limx→0 ((cos(7x) -1)/x) =Limx→0 ((7cos(7x) -1)/7x)
=7Limx→0 ((cos(7x) -1)/7x) =7(0) = 0
So, we can see how helpful it is to have these facts, and how simple they make math for math doers..
Now, let’s move on to differentiating trig functions, and we will see why these limit examples are helpful.
Differentiate the following. sin(x)
To do this, we are going to have to throw it back and use the definition of a derivative.
d/dx(sin(x)) = limh→0 (sin(x+h) - sin(x))/h
We have to simplify this further in order to be able to evaluate the limit.
If you recall we have a sum and difference formula for sin:
sin(a + b)= sina*cosb + cosa*sinb
So if we look back at our limit we have the same thing.
So, we can go ahead and write this out in place of our sin(x+h).
= limh→0 (sin(x)cos(h)+ cos(x)(sin(h) - sin(x))/h
The reason we are doing this, is to try and get things to cancel,
or to see if we can manipulate it enough to get a 1 or 0 by getting
our terms to look like our facts that we discussed earlier.
Now, let’s see if there is anything I can do to work towards getting my terms to look
like the facts that we have.
Now, because I am subtracting sin(x) it is an associative property.
So, what I’m going to do is move my -sin(x) and group it with my (sin(x)cos(h), and you’ll see why.
= limh→0 (sin(x)(cos(h)-1)+ cos(x)(sin(h)/h
Now, you can see that I haven’t actually changed anything,
because if I multiply (sin(x)(cos(h)-1) out then we are back where we started.
But what I have done is made it to where I can start to see the facts that we have.
The reason why I moved -sin(x) over with this group of terms, is because I need a -1 with my cos(h).
I wouldn’t be able to group this -sin(x) with this cos(x)sin(h), because I wouldn't have a -1
because sin(x) and sin(h) are not the same thing.
I want an h right here, so I have h/h
So it'll look like one of the facts.
Remember, like in our last problem, the variable,
and the coefficient in front of the variable need to be the same thing
for it to look like my fact.
What I’m going to do next is break up the two groups of terms that are being added together,
and take their limits separately.
= limh→0 (sin(x)(cos(h)-1)/h+ limh→0 cos(x)(sin(h)/h
Now, we can start to see how these two things are starting to look like the facts we have.
This (cos(h)-1)/h and (sin(h)/h, but we still have this pesky sin(x), and cos(x)
that are being multiplied.
But, since they are being multiplied, they are constants.
Which means that we can move them to the other side of our limit.
Which will allow us to evaluate everything to the right of our limit.
=(sin(x) limh→0 (cos(h)-1)/h + cos(x)limh→0 (sin(h)/h
Now, we have our facts that we can now evaluate.
We know that (cos(h)-1)/h is equal to zero, and (sin(h)/h is equal to one.
So, now we are left with:
d/dx(sin(x))=sin(x)(0) + cos(x)(1)
Well, sin(x)(0) is just zero and cos(x) times 1 is just cos(x).
That means that we're left with d/dx
(sin(x))=cos(x) as our answer.
That's our answer.
To differentiate cosine you would use a very similar method.
The only difference is you will need to use a different trig formula.
Once you differentiate you will get d/dx (cos(x))=-sin(x).
Once you know the derivative for sine and cosine the other four are pretty straightforward
The last trig function I’m going to differentiate for you is tangent. I will list all the other
3 trig functions and their derivatives, then we will work on some examples.
For tangent, we know that tan(x)=sin(x)/cos(x).
This is great, because we know what the derivative of sine and cosine are.
So, let’s work with what we know..
Which is really all we can do.
So, we are going to need to use the quotient rule here.
d/dx(tan(x))=d/dx(sin(x)/cos(x)) d/dx(sin(x)/cos(x))= ((sin(x))’cos(x) - sin(x)(cos(x))’)/(cos(x))2
= ((cos(x)cos(x) - sin(x)(-sin(x))/(cos(x))2 = (cos2(x) + sin2(x))/cos2(x)
Alright, once we get to this point, recall that cos2(x) + sin2(x)=1.
So, then we have:
=1/cos2(x)
And recall the definition of secant is 1/cos(x).
So, then we have our answer:
d/dx (tan(x)=sec2(x)
As I promised, here are the remaining 3 trig functions
and their derivatives for us to reference as we work on our next two examples.
Differentiate the following function.
Z(t)=4sin(t) + 7csc(t)
This problem is fairly simple.
All we need to do is recall the derivative for each of our functions.
The derivative of sin(t) is cos(t) and the derivative of csc(t) is -csc(t)(cot(t)).
Z’(t)=4cos(t)-7csc(t)cot(t)
That one was pretty simple.
Let’s try one that is a little more scary looking.
F(n)=sin(n)/(4-3cos(n))
To start, we are going to need to apply the quotient rule, then we will go from there.
F’(n)=((sin(n))’(4-3cos(n))-sin(n)(4-3cos(n))’/(4-3cos(n))2 F’(n)=((cos(n))(4-3cos(n))-sin(n)(3sin(n))/(4-3cos(n))2
Be careful to notice that when differentiating our cosine here, we already have a negative sign,
so it becomes -sin that's going to make it a positive sin.
Now from here, we can perform our operations in the numerator.
=(4cos(n)(-3cos2(n))-3sin2(n)/(4-3cos(n))2
It may seem at this point that nothing more can be done,
but there is further simplification that we can do.
Remember cos2(n)+sin2(n)=1, so to get that we can factor out a negative 3
and we'll have that same thing.
=(4cos(n)-3(cos2(n))+sin2(n)/(4-3cos(n))2
F’(n)=(4cos(n)-3)/(4-3cos(n))2
The hardest part about derivatives of trig functions, is having to recall all the formulas
and facts that you’ve forgotten from trigonometry, and pre-calculus.
So, be sure to go back, and get familiar with all that you’ve forgotten.
I hope this video was helpful.
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See you next time!