Derivatives of Trig Functions – Not So Trig(ky) Anymore
Derivatives of Trigonometric Functions
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.
First, we will take a look at two facts that will make your life simpler as you differentiate trig functions.
Limθ→0 (sin?/?) = 1
Limθ→0 ((cos?-1)/?) = 0
Being able to spot these in a function can help to make your differentiating much easier.
Before we jump into differentiation let’s take a look at a few limit problems and use our facts to help us solve them.
Well, we can move this constant to the front. So, then we have.
We know that the Lim?→0 (sin?/?) is equal to 1, because of our fact. So, what we have left is
1/24 (1). Which is 1/24.
Let’s look a more complicated one:
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.
Now, once we get to here, we have to think of ways to make this look like the fact that we have.
Well, we can multiply the whole thing by a/a, since a/a is really just 1 and doesn’t change anything.
So, then we have:
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.
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 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.
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, 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 needed to have a minus 1 with my cos(h). I wouldn’t be able to group it with cos(x)(sin(h), because for starters sin(x) and sin(h) are not the same thing, and because I need cos(h) over h. The variable, and the coefficient in front of the variable have to be the same for me to have 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 see that we have two of our facts. We have (cos(h)-1)/h and (sin(h)/h, but we still have this pesky sin, and cos that are being multiplied. Well, 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).
And we have d/dx (sin(x))=cos(x) as 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 straight forward.
The last trig function I’m going to differentiate for you is tan(x).I will list out 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). Which 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(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
= (cos^2(x) + sin^2(x))/cos^2(x)
Alright, once we get to this point, recall that cos2(x) + sin2(x)=1. So, then we have:
And recall the definition of secant is 1/cos(x).
So, then we have our answer:
As I promised, here are the remaining three 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)).
That one was pretty simple.
Let’s try one that is a little more scary looking.
To start, we are going to need to apply the quotient rule, then we will go from there.
Be careful to notice that when differentiating our cosine the negative signs will cancel to give us a positive. So, now from here, we can perform our operations in the numerator.
It may seem at this point that nothing more can be done, but there is further simplification that we can do. Remember cos^2(n)+sin^2(n)=1, so to get that we can factor out a negative 3 from our cos^2 and sin^2.
Now, cos^2(n)+sin^2(n) is equal to 1, so our answer is:
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. For further help, check out more of our videos by subscribing below.
See you next time!