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## Calculus 1

### Course: Calculus 1>Unit 2

Lesson 12: Derivatives of tan(x), cot(x), sec(x), and csc(x)

# Derivatives of sec(x) and csc(x)

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?
• 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!
• 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?
• For AP Calculus BC, is it expected that these 6 identities be memorized?
• In every calculus sequence I've seen, it was expected that you memorize the identities . Good luck!
• Use definition of derivative to show that
d÷dx(secx)=secx tanx
• 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
• 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?
• 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.
• 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!
• 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.
• is (cscx)' also equal to -csc^2x*cosx ?
• 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
• 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?