Derivatives of Trigonometric Functions
To determine the derivative of trigonometric functions I am going to start out by graphing
one full cycle of the sine curve. You should be familiar with this function from your studies
of precalculus or math analysis. So from 0 to 2 pi the graph looks like this.
It has maximum of one and a minimum of negative one,
So that's how the graph of sine of x looks like.
Now what I am going to do is I am going to plot a graph with the inputs corresponding
to the radian measure and my output to correspond to the measure of the slope of the tangent line at that given
input. Let's start plotting.
So as you notice the tangent line initially is positive specifically it has a value of one.
so my first point is (0, 1). So now let's see what happens when you graph more points.
So initially the tangent line is still positive, so my output should still be giving me positive values
but then slowly the slope of the tangent line attains a value of zero.
It attains a value of zero at pi over 2.
So at this point the slope of the tangent line becomes negative,
so I start getting negative values for the output,
and then as I head towards pi, and head towards 3 pi over 2
the slope of the tangent line is once again slowly achieving a value of zero.
Until it achieves it at 3 pi over 2,
and then at this point the slope of the tangent line becomes positive once more
so the values of the output are positive.
Ok so there you go there's one full cycle.
So now I am looking at the blue graph it looks awfully familiar and indeed this is the cosine function.
So it turns out that the derivative of sine of x is actually cosine of x, pretty neat huh?
Ok so now that we know how to find the derivative of sine of x
let's try to figure out what the derivative of cosine of x is equal to.
Just like the previous example I am going to graph one full cycle of the cosine curve.
So it has a maximum of one and a minimum of negative 1 so this is one full cycle of cosine of x.
and again just like the previous example
I am going to graph the value of the tangent line at every single x value.
Ok so let's do just that
and as you notice our first point here the value of the slope of the tangent line is equal to zero.
So our first point is actually (0,0).
So as we generate points the slope initially is negative,
so we should expect to get a negative output for our graph
and then as it transitions to pi over 2 it begins to slowly slowly attain a value of zero.
Until the value of the tangent line is actually zero at pi.
So here the tangent line is zero
and then from here the tangent line becomes positive so the values become positive.
and then as it approaches 3 pi over 2
the slope of the tangent line slowly attains a value of zero.
and finally achieves value of zero at 2 pi.
It's hard to see at first but the red graph is actually an inverted sine curve.
So it turns out that the derivative of cosine is actually the negative of sine, preety cool.
Ok so now that we know what the derivative of sine
and cosine is we can find the derivative of other trigonometric functions let's find the derivative
of tangent. Find the derivative of f of x equals tangent
of x. First we need to rewrite this function in
terms of sine and cosine remember tangent of x can be rewritten as sine over cosine in other words
a quotient. We can use the quotient rule to find the derivative.
So rewriting the expression as sine over cosine
and letting f of x to be equal to sine of x and g of x to be equal to cosine of x.
we can use the quotient rule as follows,
the derivative of R of x is going to be equal to
cosine of x times the dervative of sine of x minus sine of x times the dervative of cosine of x
all over cosine of x squared.
So applying appropriate differentiation rules we get that the derivative of sine is cosine,
and the derivative of cosine is negative sine.
simplifying the expression gives us the following.
Now here we can rewrite the expression cosine squared plus sine squared of x.
Using one of the trigonometric identities back from precalculus class.
Remembering the identity cosine squared of x plus sine sqaured of x equals 1,
we can substitute the numerator with just 1.
So the derivative of R of x is going to be equal to one over cosine squared of x,
and remembering that the reciprocal of cosine is secant
and the reciprocal of secant is 1 over cosine we can rewrite the expression as
secant squared of x.
So in the end the derivative of tangent is equal to secant squared of x.
So with this we add another trigonometric derivative to our list. Let's find some more.
Find the derivative of cosecant of x.
Ok so remembering that the reciprocal of sine is cosecant and the reciprocal of cosecant is 1 over sine,
we can rewrite this expression as R of x equals 1 over sine of x.
Here we let f of x to be equal to 1 and g of x to be equal to sine of x.
Using the quotient rule we have that the derivative of R of x is going to be equal to sine of x,
times the derivative of 1 minus 1 times the derivative of sine of x all over sine of x squared.
Applying appropriate derivatives and simplifying the expression gives us the following,
now here what we want to do is we want to rewrite this expression
so we can break up the denominator as follows to give us the expression negative times
one over sine of x times cosine of x over sine of x.
Now one over sine of x is the same thing as cosecant of x so we rewrite that,
and cosine over sine is just another way of writing cotangent of x.
So the final derivative of R of x is equal to negative cosecant times cotangent of x.
By using the quotient rule you can find the derivative of the reciprocal trigonometric function,
and I encourage you to derive the derivative for secant and cotangent in the same manner.
Check your answer with the following table.
Here we have the derivative of all six trigonometric functions.
You really don't need to memorize all the derivatives only that of sine and cosine.
As we just saw in the previous problems we can derive all the other derivatives by applying the quotient rule
But if you really want to memorize the table here is a tip.
Notice that the minus signs go with the derivatives of the "co" functions
Specifically cosine, cosecant, and cotangent.
Ok let's finish of this video with a final example.
Find the derivative of R of theta equals cosecant of theta times the quantity of theta plus cotangent of theta.
We have a product of functions so this is a perfect candidate for the product rule.
I am going to let f of theta to be equal to cosecant of theta
and g of theta to be equal to theta plus cotangent of theta.
Applying the product rule we have that the derivative of R of theta is going to be equal to cosecant of theta
times the derivative of theta plus cotangent of theta plus the quantity of theta plus cotangent of theta times
the derivative of cosecant of theta.
Ok so simplifying the expression and distributing gives us the following,
and further distributing gives us the following expression for the final derivative of the function.
Alternatively we can rewrite this expression as follows we can factor out a cosecant of theta
and using one of the other trigonometric identities ok again you should of learned this from precalculus class,
one plus cotangent squared of theta equals cosecant sqaured of theta.
So we can substitute this expression for cosecant squared of theta which gives us the following.
Simplifying it further the one's cancel out and we distribute the negative sign,
and then here we collect like terms,
and then here we factor out a negative and a cotangent of theta.
So an alternative form for the answer for the derivative of R of theta is equal to negative cosecant of theta
times cotangent of theta times the quantity 2 times cotangent of theta plus theta.
So in our next video we are going to learn the chain rule
and with this final derivative rule we could essentially find any derivative in calculus.