Let's explore the derivatives of sec(x) and csc(x) by expressing them as 1/cos(x) and 1/sin(x), respectively, and applying the quotient rule. We discover that the derivative of sec(x) can be written as sin(x)/cos²(x) or tan(x)sec(x), and the derivative of csc(x) can be expressed as -cos(x)/sin²(x) or -cot(x)csc(x).
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- How can you prove the sex and cscx derivatives using the power rule? Because you can't take them as cos^-1x or sin^-1x, can you?(6 votes)
- You can prove the sec x and cosec x derivatives using a combination of the power rule and the chain rule (which you will learn later).
Essentially what the chain rule says is that
d/dx(f(g(x)) = d/dg(x) (f(g(x)) * d/dx (g(x))
When you have sec x = (cos x)^-1 or cosec x = (sin x)^-1,
you have it in the form f(g(x)) where f(x) = x^-1 and g(x) = cos x or sin x.
Below is the working for how to derive the derivatives of sec x using this:
d/dx (sec x)
= d/dx ((cosx)^-1)
= -1 * (cos x)^-2 * d/dx (cos x)
= -1 * (cos x)^2 * (-sin x)
= sin x/(cosx)^2
= sec x * tan x
The same process can be repeated for cosec x and as can be clearly seen, the results are the same as what Sal gets.
Hope I helped!(11 votes)
- So, just by looking at the graphs, it is obvious that d/dx[sin(x)]=cos(x), & d/dx[cos(x)]=-sin(x). However, I'm just curious about the mathematical proof of it. Can somebody please redirect me to a video with the proof?(4 votes)
- You can check out a really great proof here: https://www.wyzant.com/resources/lessons/math/calculus/derivative_proofs/sinx(5 votes)
- For AP Calculus BC, is it expected that these 6 identities be memorized?(2 votes)
- Use definition of derivative to show that
d÷dx(secx)=secx tanx(2 votes)
- 𝑓(𝑥) = sec 𝑥 = 1∕cos 𝑥 ⇒
⇒ 𝑓 '(𝑥) = lim(ℎ → 0) [(1∕cos(𝑥 + ℎ) − 1∕cos 𝑥)∕ℎ] =
= lim(ℎ → 0) [(cos 𝑥 − cos(𝑥 + ℎ))∕(ℎ cos 𝑥 cos(𝑥 + ℎ))] =
= lim(ℎ → 0) [(cos 𝑥 − cos 𝑥 cos ℎ + sin 𝑥 sin ℎ)∕(ℎ cos 𝑥 cos(𝑥 + ℎ))] =
= lim(ℎ → 0) [(1 − cos ℎ + tan 𝑥 sin ℎ)∕(ℎ cos(𝑥 + ℎ))] =
= lim(ℎ → 0) [(1 − cos ℎ)(1 + cos ℎ)∕(ℎ(1 + cos ℎ) cos(𝑥 + ℎ)) +
+ tan 𝑥 sin ℎ∕(ℎ cos(𝑥 + ℎ))] =
= lim(ℎ → 0) [sin²ℎ∕(ℎ(1 + cos ℎ) cos(𝑥 + ℎ)) + tan 𝑥 sin ℎ∕(ℎ cos(𝑥 + ℎ))] =
= lim(ℎ → 0) [(sin ℎ∕ℎ) ∙ (sin ℎ∕((1 + cos ℎ) cos(𝑥 + ℎ)) + tan 𝑥∕cos(𝑥 + ℎ)] =
= 1 ∙ (0∕(2 cos 𝑥) + tan 𝑥∕cos 𝑥) =
= tan 𝑥 sec 𝑥(7 votes)
- I always had a hard time trying to not mix up the concepts of the tangent function with concept of the tangent line. I have used a good deal of hours to try to understand the concepts and establish a relation between them and somehow it still isn't very easy for me to not mix it up sometimes.
Now I am presented with the function sec x, that I have missed from the geometry playlist, apparently. After getting this far in the calculus playlist I imediatly though of the sec x function having something to do with secant line but I fail to establish a clear relationship. I went to check the definition of sec x and I now know that it is the inverse of the cos x function but I can't help but try to make a relation with secant line without being able to in a satisfying manner. Any hint on how to think about this?(2 votes)
- It's unfortunate that "secant" and "tangent" serve double purposes in mathematics, but we're stuck with them now. This isn't really a calculus matter, it's trigonometry. There's a neat interactive diagram relating the six functions - sine, cosine, tangent, cotangent, secant, and cosecant - on this page:
In the "Appendix: All trig ratios in the unit circle" section. (About 2/5 down the page, just before the Q&A's.
You may be even able to see how tangent (the function) and a tangent line are related. I can see no obvious reason why the secant function is so named though.(6 votes)
- I'm a little confused overall because I thought the derivative of sec(u) was sec(u)tan(u)*du/dx. Is there a theorem or a law for the derivative of these trig functions? Thank you to whoever can answer!(2 votes)
- You didn't make any mistake. But I think the problem is that you have assumed u as a function of x. That's why by the chain rule you would get the du/dx part. In the video, Sal is only considering one variable x and has taken the derivative with respect to x. He has not taken a function of x. If he were to consider a function of x and then take the derivative of that function with respect to x then his answer would look similar to yours.(3 votes)
- is (cscx)' also equal to -csc^2x*cosx ?(2 votes)
- If f(x) = csc(x) then f'(x) = -cot(x)csc(x) = (-cos(x)/sin(x))(1/sin(x)) = -cos(x)/((sin(x))^2) = -cos(x)*(csc(x)^2) so yes(2 votes)
- What means d/dx? That we need to find the derivative?
Is there a difference between dx/d and d/dx ?
Is dx/d suppose mean that we need to find integral?(2 votes)
- Is it more efficient or beneficial to memorize the derivatives of sec(x), csc(x), tan(x) and cot(x) OR to spend extra time working through the proofs that Sal has just done for us here?(1 vote)
- Personally, I'd advise you to practice problems with them. That way, you'll get used to using the derivatives and eventually, you'll remember them. Proofs are also good to know though, as they can help you derive the derivative incase you forget(3 votes)
- [Voiceover] In a previous video we used the quotient rule in order to find the derivatives of tangent of x and cotangnet of x. And what I what to do in this video is to keep going and find the derivatives of secant of x and cosecant of x. So let's start with secant of x. The derivative with respect to x of secant of x. Well, secant of x is the same thing as so we're going to find the derivative with respect to x of secant of x is the same thing as one over, one over the cosine of x. And that's just the definition of secant. And there's multiple ways you could do this. When you learn the chain rule, that actually might be a more natural thing to use to evaluate the derivative here. But we know the quotient rule, so we will apply the quotient rule here. And it's no coincidence that you get to the same answer. The quotient rule actually can be derived based on the chain rule and the product rule. But I won't keep going into that. Let's just apply the quotient rule right over here. So this derivative is going to be equal to, it's going to be equal to the derivative of the top. Well, what's the derivative of one with respect to x? Well, that's just zero. Times the function on the bottom. So, times cosine of x. Cosine of x. Minus, minus the function on the top. Well, that's just one. Times the derivative on the bottom. Well, the derivative on the bottom is, the derivative of cosine of x is negative sine of x. So we could put the sine of x there. But it's negative sine of x, so you have a minus and it'll be a negative, so we can just make that a positive. And then all of that over the function on the bottom squared. So, cosine of x, squared. And so zero times cosine of x, that is just zero. And so all we are left with is sine of x over cosine of x squared. And there's multiple ways that you could rewrite this if you like. You could say that this is same thing as sine of x over cosine of x times one over cosine of x. And of course this is tangent of x, times secant of x. Secant of x. So you could say derivative of secant of x is sine of x over cosine-squared of x. Or it is tangent of x times the secant of x. So now let's do cosecant. So the derivative with respect to x of cosecant of x. Well, that's the same thing as the derivative with respect to x of one over sine of x. Cosecant is one over sine of x. I remember that because you think it's cosecant. Maybe it's the reciprocal of cosine, but it's not. It's the opposite of what you would expect. Cosine's reciprocal isn't cosecant, it is secant. Once again, opposite of what you would expect. That starts with an s, this starts with a c. That starts with a c, that starts with an s. It's just way it happened to be defined. But anyway, let's just evaluate this. Once again, we'll do the quotient rule, but you could also do this using the chain rule. So it's going to be the derivative of the expression on top, which is zero, times the expression on the bottom, which is sine of x. Sine of x. Minus the expression on top, which is just one. Times the derivative of the expression on the bottom, which is cosine of x. All of that over the expression on the bottom squared. Sine-squared of x. That's zero. So we get negative cosine of x over sine-squared of x. So that's one way to think about it. Or if you like, you could do this, the same thing we did over here, this is the same thing as negative cosine of x over sine of x, times one over sine of x. And this is negative cotangent of x. Negative cotangent of x, times, maybe I'll write it this way, times one over sine of x is cosecant of x. Cosecant of x. So, which ever one you find more useful.