BAM!!!
Mr. Tarrou.
I have explained out the derivative of sine is equal to cosine, or at least stated it,
and I explained with a graphical demonstration why the derivative with respect to x of cosine
of x is equal to the negative sine of x.
Now we have got four other trig functions of course.
And we have to know the derivatives of those as well.
So the derivative with respect to x of the secant of x is equal to the secant of x times
the tangent of x.
The derivative with respect to x of cosecant of x is equal to negative cosecant of x times
the cotangent of x.
And so on and so on.
I am sure you can read this.
What we are going to do in this video is prove two of these four statements.
I am sure in your homework you will be required to prove, well hopefully I am not just giving
you the answer to one of your homework problem, but you will have and should be able to prove
the other two.
Whatever other two that I don't do.
So we are going to take a look at the derivative with respect to x of the cosecant of x.
Now we are going to look at this the same way that we did our proofs in Trigonometry.
That is sine and cosine are building blocks of the other four trig functions.
Like tangent is sine over cosine.
And cotangent is cosine over sine.
Well cosecant is one over sine.
So we have the derivative with respect to x of the cosecant of x.
Well I just told you want the answer was, right here.
But where does this come from?
Why is there all of a sudden two trig functions?
Well to figure that out we are going to write this in terms of sine and cosine.
That is going to be now the derivative with respect to x of one over the sine of x.
Now one over the sine of x.
We know how to...
We know the derivative of sine is cosine.
And I showed this second statement graphically and showed the two together to help you accept
it as true.
If I want to find the derivative of this, well I have one divided by the sine of x.
So that is means that we need the quotient rule since at this point that is all we have
covered.
So the quotient rule says that we need to take the denominator, so that is going to
be the sine of x times the derivative with respect to x of one minus... ok so it is the
denominator times the derivative of the numerator... and then it is the derivative of the denominator
times the numerator... all over the denominator squared.
So it is all over sine squared x.
So the derivative of a constant is equal to zero.
So this first term in the numerator is just going to be zero.. minus... so we have minus
the derivative of sine is equal to cosine...
so the derivative with respect with the sine of x is equal to cosine of x times one over
sine squared x.
Well that does not look like this.
What are we doing again?
Cosecant.
It does not look like this, negative cosecant of x times cotangent of x.
That is an issue when you derivatives especially with trig functions.
You might have a correct answer, this is the derivative of the cosecant function, but darn
it when I go in the back of the book and I look at the answer it does not match up.
Or when I look at the formula or the rule that I just gave you it does not match.
We are going to manipulate this a little bit.
This has a squared trig function in the bottom and these all have exponents of one.
It would be there.
So we are going to take this denominator and pull it apart and rewrite this as negative
cosine x over the sine of x and I am going to show this as a completely different multiplication.
Because here I have a term of two factors.
So I have negative cosine x over sine x times...
well.. negative cosine x times one is going to equal negative cosine x.
I have to multiply by one because I don't want the value to change.
We have a sine squared in the denominator, so that need to be sine x.
So now I can see a little bit better about where this is coming from.
Except mine is going to be in a different order, but multiplication is commutative.
Cosine of sine is cotangent, so this is negative cotangent of x.
And one over sine x is cosecant x.
That is the end of my proof of this individual rule right here.
BAM!
We are going to do one more example and that will be the end of this video.
For our second and last example we have the derivative with the respect to x of the cotangent
of x.
Why is that equal to negative cosecant squared x.
Well the building block of cotangent are cosine over sine.
So this is equal to the derivative with respect to x of the cosine of x over the sine of x.
That means that we are going to need the quotient rule again.
That is going to be equal to the denominator so sine of x times the derivative of the numerator
which is going to be...
Ok, let's write it out.
The derivative with respect to x of cosine x minus... ok so it was denominator times
the derivative of the numerator... now it is the derivative of the denominator times
the numerator... all over the denominator squared.
Ok, so we have sine x, the derivative of cosine is negative sine x, minus the derivative of
sine which is cosine x times cosine x, all over sine squared x.
Now sine times sine is sine squared, but there is a negative there so negative sine squared
x minus cosine times cosine is cosine squared x... so we have cosine squared x all over
sine squared x.
It is just like doing a trig proof.
We have to keep algebraically manipulating things along, but we are having to keep our
mind our goal.
I am trying to get this equal to negative cosecant squared x to finish my proof.
Never thought you would be doing those again maybe after you did them in trig.
They do come back in Calculus.
So how am I going to get this to look like this.
Well cosecant is one over sine.
Well so far I have sine and cosine.
I need to get cosine squared out and replace it with sine if I have any hopes of getting
this negative cosecant squared to show up.
Well, negative sine squared x minus... something over sine squared x.
Pythagorean Theorem, you had three of those to memorize in Trigonometry.
They will come back and you will have to use them again in Calculus.
The first one was sine squared x plus cosine squared x equals one.
So that means that if sine squared plus cosine squared equals one, then cosine squared when
I get that sine squared and move it over...
Cosine squared is equal to 1 minus sine squared x.
So now if I take this negative sign and I distribute it through the parenthesis...
Is that going to get me any closer?
I had a little brain fade there.
So of course it is.
So we are going to get...
Running out of room here.
We have negative one.
Now we have minus sine squared x and a negative times a negative is positive.
So we have a negative sine squared x and will be with this distribution a positive sine
squared x, that means that these are going to cancel out.
So I was trying to do a little bit in my head there because I was running out of room.
And just whatever.
Spaced out.
So we have negative one over sine squared x.
And um, one over sine squared is equal to cosecant squared.
So we have negative cosecant squared x and the end of my second proof is done.
Now you go try and finish the proofs of the last ones.
BAM!!!
I am Mr. Tarrou.
Go Do Your Homework:D