yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Trig functions differentiation | Derivative rules | AP Calculus AB | Khan Academy


4m read
·Nov 11, 2024

So let's say that we have ( y ) is equal to the secant of (\frac{3\pi}{2} - x), and what we want to do is we want to figure out what (\frac{dy}{dx}) is, the derivative of ( y ) with respect to ( x ) at ( x = \frac{\pi}{4} ).

Like always, pause this video and see if you could figure it out. Well, as you can see here, we have a composite function; we're taking the secant not just of ( x ), but you could view this as of another expression that I guess you could define or as of another function.

So, for example, if we call this right over here ( u(x) ), so let's do that. If we say ( u(x) ) is equal to (\frac{3\pi}{2} - x), we could also figure out ( u' ) of ( x ) is going to be equal to the derivative of (\frac{3\pi}{2}); that's just going to be zero. The derivative of (-x) is going to be (-1), and you could just view that as a power rule; it's ( 1 \cdot -1 \cdot x^{0} ), which is just one.

So there you go! We could view this as the derivative of secant with respect to ( u(x) ), and when we take the derivative, the derivative of secant with respect to ( u(x) ) times the derivative of ( u ) with respect to ( x ).

You might say, "Well, what about the derivative of secant?" Well, in other videos, we actually prove it out, and you could actually re-derive it. Secant is just ( \frac{1}{\cos(x)} ), so it comes straight out of the chain rule.

So in other videos, we proved that the derivative of the secant of ( x ) is equal to (\sec(x) \tan(x)). So if we're trying to find the derivative of ( y ) with respect to ( x ), well, it's going to be the derivative with respect to ( u(x) ) times the derivative of ( u ) with respect to ( x ).

So let's do that. The derivative of secant with respect to ( u(x) ) well, instead of seeing an ( x ) everywhere, you're going to see a ( u(x) ) everywhere. So this is going to be (\sec(u(x)) \tan(u(x))).

I don't have to write ( u(x) ); I could write (\frac{3\pi}{2} - x), but I'll write ( u(x) ) right over here just to really visualize what we're doing: (\sec(u(x)) \tan(u(x))).

So that's the derivative of secant with respect to ( u(x) ), and then the chain rule tells us it's going to be that times ( u' ). ( u' ) of ( x ) we already figured out is (-1), so I could write (\sec(u(x)) \tan(u(x)) \cdot u' ) where ( u' ) of ( x ) we already figured out is (-1).

Now, we want to evaluate ( \frac{dy}{dx} ) at ( x = \frac{\pi}{4} ). So when that is equal to ( \frac{pi}{4} ), let's see. This is going to be equal to (\sec\left(\frac{3\pi}{2} - \frac{\pi}{4}\right)\tan\left(\frac{3\pi}{2} - \frac{\pi}{4}\right) \cdot -1).

So if you have a common denominator, that is (\frac{6\pi}{4} - \frac{\pi}{4} = \frac{5\pi}{4}). So it's (\sec\left(\frac{5\pi}{4}\right) \tan\left(\frac{5\pi}{4}\right) \cdot -1).

Now, what is (\sec\left(\frac{5\pi}{4}\right)) and (\tan\left(\frac{5\pi}{4}\right))? Well, I don't have that memorized, but let's actually draw a unit circle, and we should be able to figure out what that is.

So a unit circle... I try to hand-draw it as best as I can; please forgive me that this circle does not look really like a circle! Alright, okay, so let me just remember my angles. In my brain, I sometimes convert into degrees. (\frac{\pi}{4}) is (45°), this is (\frac{\pi}{2}), this is (\frac{3\pi}{4}), this is (\frac{4\pi}{4}), this is (\frac{5\pi}{4}), lands you right over there.

So if you wanted to see where this intersects the unit circle, this is at the point where your ( x )-coordinate is (-\frac{\sqrt{2}}{2}) and your ( y )-coordinate is (-\frac{\sqrt{2}}{2}).

If you're wondering how I got that, I encourage you to review the unit circle and some of the standard angles around the unit circle; you'll see that in the trigonometry section of Khan Academy. But this is enough for us because the sine is the ( y )-coordinate. So (\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}).

So this is (-\frac{\sqrt{2}}{2}), and then the cosine is the ( x )-coordinate, which is also (-\frac{\sqrt{2}}{2}), but it's going to be that squared: (\left(-\frac{\sqrt{2}}{2}\right)).

So if we square this, it's going to become positive, and then (\left(-\frac{\sqrt{2}}{2}\right)^{2} = \frac{2}{4} = \frac{1}{2}), so this is the denominator.

In the numerator, the negative cancels out with that negative, and so we are left with—and we deserve a little bit of a drum roll— that we are left with (\frac{-\frac{\sqrt{2}}{2}}{\frac{1}{2}}).

Well, that's the same thing as multiplying by (2), so we are left with (\sqrt{2}). This is the slope of the tangent line to the graph of ( y ) is equal to this when ( x ) is equal to (\frac{\pi}{4}). Pretty exciting!

More Articles

View All
Magnetic forces | Forces at a distance | Middle school physics | Khan Academy
Let’s talk about magnets and magnetic forces. Magnets are these neat objects that are able to attract metals like iron. Magnets are used in all sorts of things, from holding paper on your refrigerator to computers to compasses. So, magnets can be used to …
Adam Brown on how to be resilient during a time of high stress and anxiety | Homeroom with Sal
Hi everyone, welcome to the daily homeroom live stream. Sal here from Khan Academy. For those of you who are wondering what this is, this live stream is something we started as soon as we saw schools starting to get closed around the world. Because we saw…
Not The Confederate Flag?
This is not the confederate national flag: When the United States split in twain during the Civil War, this was the first flag her rebel half used: The Bonnie Blue, which she copied from the Republic of West Florida. No, really. This country existed: a bo…
BEST Images of the WEEK! ... IMG! #28
When your house breaks, fix it. And a private toilet! It’s episode 28 of IMG. Do you like cake? Do you like tacos? Well, get yourself a Taco Bell cake covered in say cheese, then say arson. Just don’t burn down the melting stairs. Wieners for kids! But z…
Why YOU Need To Invest in PSYCHEDELICS | Ask Mr. Wonderful #14 Kevin O'Leary
[Music] All right everybody, back for another episode of “Ask Mr. Wonderful.” Here today with my recently acquired 1969 Telecaster. Telecasters are very unforgiving guitars. Not that I want to get sidetracked here, but I just thought maybe a couple of lit…
Homeroom with Sal & Congresswoman Karen Bass - Wednesday, August 26
Hi everyone, Sal Khan here. Welcome to this Homeroom live stream. As always, I’m very excited about the conversation we’re going to have with our guest today, Representative Karen Bass. But before we get to that, I’ll give my standard announcements. Fir…