Many students can feel relatively comfortable with the content of their algebra or calculus courses until trig functions are thrown in the mix.
In this video, we are going to discuss the derivatives of each of the main trig functions, the derivatives of the inverse trig functions,
and a short chain rule tutorial, to show you that there is a nothing to fear.
By now you probably know that, the derivative of, the natural log of x, is 1 over x.
This is typically learned from repetition and memorization.
Because the derivatives of trigonometric functions are similar in this regard,
the purpose of this video will be to give you a familiarization with each of the trig functions’ derivatives.
Let’s begin with the sine function.
Believe it or not, the derivative of, sine of x, is cosine of x.
The derivative of cosine of x, on the other hand, is equal to, negative sine of x.
It’s very important to remember the negative sign.
It’s worth noting that these derivatives can be proven using the definition of the derivative and the sum and difference identities.
Since these proofs are quite lengthy, we will not cover them in this video.
Moving along, the derivative of, tangent of x, is secant squared of x.
Recall that secant is equivalent to, 1 over cosine of x.
Before we go on to discuss the derivatives of the reciprocal functions secant, cosecant, and cotangent, let’s do a couple examples.
Find the derivative of, f of x is equal to, cosine of x, plus tangent of x.
Since this function simply involves the addition of cosine and tangent, we can handle one term at a time.
As we mentioned, the derivative of cosine of x, is negative sine of x, and the derivative of tangent of x, is secant squared of x.
So f prime of x is equal to, negative sine of x, plus secant squared of x.
Let’s try another problem.
Find the fourth derivative of, g of x is equal to, sine of x.
This may sound like an absurd problem, but the process is relatively simple.
In order to find the fourth derivative, we must start by finding the first derivative, and then the second, and so on.
Well, the derivative of sine of x, is cosine of x, so the first derivative, g prime of x, is equal to cosine of x.
Taking the derivative of this will give us the second derivative.
Since the derivative of cosine is negative sine, we now get, g double prime of x is equal to, negative sine of x.
Let’s take the derivative again to get g triple prime.
This time, we are basically taking the derivative of sine again but carrying a negative sign along with it.
We get, g triple prime of x is equal to, negative cosine of x.
For the final leg of this problem, we will take the derivative once more.
Since we have negative cosine, we are going to write the negated form of cosine’s derivative.
This gives us, the fourth derivative of g is equal to, sine of x.
As we can see from this result, the sine and cosine functions form an interesting cycle with each other each time you take their derivative.
In the same way that the fourth derivative of sine is itself, the fourth derivative of cosine is cosine!
Let's now continue with the derivatives of the reciprocal trig functions, starting with cosecant.
The derivative of cosecant of x is, negative cosecant of x, times cotangent of x.
The derivative of secant of x is, secant of x, times tangent of x.
And finally, the derivative of cotangent of x is, negative cosecant squared of x.
Take a moment now to write down each of these derivatives.
In fact, if you dedicate just a few minutes to writing them again and again, you’ll start to become familiar with each one,
which will give you a huge advantage on memorizing them.
Even though memorization is rarely an exciting thing to do, you’ll find that it will really help you down the road in future assignments.
Pause the video now while you write out these derivatives.
Let’s now shift to derivatives of the inverse trig functions.
But before jumping in, let’s very quickly review what inverse trig functions are.
Say we have, sine of a, equals b.
In this expression, a is some angle, and b is the y-coordinate associated with that angle on the unit circle.
The inverse sine function swaps the places of a and b.
So the inverse function takes a y-coordinate, like b, on the unit circle as its argument, and returns the angle a, which corresponds to that point.
The other inverse trig functions work in the same way.
For example, cosine inverse of x returns the angle theta, which would solve the equation, cosine of theta, equals x.
Now, the derivatives of each of the inverse trig functions look considerably different from those of the regular trig functions.
Let’s check them out now.
The derivative of, sine inverse of x, is one over, the square root of, one minus x-squared.
Note that when x equals 1, the denominator becomes 0.
For this reason, the derivative of sine inverse is undefined at that point.
Next, the derivative of cosine inverse is, negative 1 over, the square root of, 1 minus x-squared.
Once again, this derivative is undefined when the denominator equals 0.
The derivative of tangent inverse is, 1 over, 1 plus x-squared.
This derivative has no possibility of the denominator equaling 0, so it is defined at all points x.
Before we go on with the other three, let’s try a couple of examples with these derivatives.
Find the derivative of, f of x is equal to, sine inverse of x, minus cosine inverse of x.
To get this derivative, we just need to handle f one term at a time.
The first term is sine inverse of x, and we know that its derivative is, 1 over, the square root of, 1 minus x-squared.
The second term is negative, so we are going to negate the derivative of the cosine inverse.
Since the second term of f prime has a double negative, we can rewrite it as a positive, then add the two terms together since they share the same denominator.
Let’s try another problem.
Find the derivative of, g of x is equal to, x times the tangent inverse of x.
Since x and tangent inverse of x are multiplied together, this is an example of a product rule problem.
To take the derivative, we write the first factor times the derivative of the second, plus the second factor times the derivative of the first.
As we have said, the derivative of tangent inverse is, 1 over, 1 plus x-squared, and of course, the derivative of x is simply 1.
So, using the product rule, we know that g prime is equal to, x times, 1 over, 1 plus x-squared, plus the tangent inverse of x.
Let’s go through the derivatives of the three remaining inverse trig functions now.
Next in line is cosecant inverse, whose derivative is, negative 1 over, the absolute value of x, times the square root of, x-squared minus 1.
This derivative is undefined whenever x equals 0 or positive or negative 1.
Perhaps unsurprisingly, the derivative of secant inverse is identical to this but with a positive sign.
Due to the same denominator, though, this derivative is also undefined at the points x equals 0, x equals 1, and x equals negative 1.
Finally, the derivative of cotangent inverse is, negative 1 over, 1 plus x-squared.
Notice that this is identical to the derivative of tangent inverse but with a negative sign.
I want to reiterate that memorization is important.
One tip that may help you remember the derivatives of the inverse trig functions is that the “co” functions (COsine, COsecant, and COtangent inverses)
each have negative derivatives.
While inverse trig functions and their derivatives don’t come up quite as frequently in Calculus I as the regular trig functions do,
you are still encouraged to familiarize yourself with them for those times when they do come up.
In fact, a solid grasp of the inverse trig derivatives will come in handy down the road in the first unit of Calculus II.
Now that we’ve discussed what each of the derivatives are for the regular and inverse trig functions, let’s briefly discuss how the chain rule can be applied to them.
For example, how would you take the derivative of, h of x is equal to, the cosine of, 2 x-squared?
Remember, the chain rule allows us to handle derivatives of “nested” functions, which have inner and outer components.
In this case, cosine is the outer function, and 2 x-squared is the inner function.
The chain rule states that we can find the derivative of composite functions like these by first handling the derivative of the outer function—
with the inner function untouched—and then multiply this result with the derivative of the inside function.
For this function then, we start by looking at the outer function, cosine, and determining its derivative.
Do you remember what it is?
The derivative of cosine is negative sine, so we are going to write, h prime of x is equal to, negative sine of, 2 x-squared,
and then multiply this by the derivative of 2 x-squared, which is 4x.
Not so bad right?
Chain rule problems involving trig functions are very similar to the chain rule problems you’ve already encountered.
Let’s work an example involving one of the inverse trig functions now.
Find the derivative of, k of x is equal to, cotangent inverse of, 8x plus 1.
The outer function, cotangent inverse, has a derivative of, negative 1 over, 1 plus x-squared, so we will start by writing that but with, 8x plus 1, in place of x.
Now, we will multiply this by the derivative of 8x plus 1, which is 8.
If you haven’t already spent a few minutes writing out each of these derivatives, go ahead and do so now.
You’ll be thanking yourself for it later during class and future assignments!
Thanks for watching, and happy studying!