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

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!