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### Course: Differential Calculus>Unit 2

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

# Derivatives of tan(x) and cot(x)

We find the derivatives of tan(x) and cot(x) by rewriting them as quotients of sin(x) and cos(x). Using the quotient rule, we determine that the derivative of tan(x) is sec^2(x) and the derivative of cot(x) is -csc^2(x). This process involves applying the Pythagorean identity to simplify final results.

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• I have always seen the derivative of tan(x) as sec^2(x) and the derivative of cot(x) as -csc^2(x). This seems to be the standard, and I have never seen it otherwise. However, Sal is using 1/cos^2(x) as the derivative of tan(x) and -1/sin^2(x) as the derivative of cot(x). He goes on to prove that the the different derivatives are actually the same, but he seems to prefer the sin/cos versions, and in the exercises, those are used. Is there a reason for this preference? Which one should I use? Is Sal's way somehow simpler, and if so, why isn't it in use elsewhere (at least as far as I could tell)?
• First, we need to review the trig functions. We know the 2 basic ones, sinx and cosx
From these 2 we built 4 more.
tanx = sinx/cosx
cotx = 1/tanx = cosx/sinx
secx = 1/cosx
cscx. = 1/sinx

As you can see, if secx= 1/cosx, then sec²x=(1/cosx)² = 1/cos²x, similarly, -csc²x = - 1/sin²x
They are equivalent, either is fine. It's personal preference but depends on how you are asked to simplify. Some consider sec²x or -csc²x are the simplified answer to what Sal has, such that textbooks will use them over the reciprocal forms.

Unless you have it memorized, you wouldn't be able to find what sec(π/3) is without turning into 1/cos(π/3). So it's more useful to have it in the sinx or cosx form if you are required to do calculations.

Although, the "standard" derivative of tanx is sec²x, a sub instructor told us to get into a habit of using tan²x + 1 instead (this comes from tan²x + 1 = sec²x identity) because it is used more often in calculus 2.
• Can someone walk me through (step by step) how to derive tan(x) via the product rule? Or correct me where I wrong:

Step one:
define the function and set up the formula
F=f(x)g(x)
F'=f ' (x)g(x) + f(x) g ' (x)
f(x) = sin(x) g(x) = [cos(x)]^(-1)

Step two:
Derive F (x). Derivative of sin(x) is cos(x) multiplied by [cos(x)]^(-1) all that PLUS sin(x) multiplied by derivative of [cos(x)]^(-1) which needs the chain rule. (is that correct?). bring the (-1) down, and subtract 1 from the exponent ... then the derivative of cos(x)
F' = cos(x)*[cos(x)]^(-1) + sin(x)*(-1){[cos(x)]^(-2)}*[-sin(x)]
= [cos(x)/cos(x)]+ [(sin(x)^2)/(cos(x)^2]
=1+tan(x)^2

Where did I go wrong?. I know I need ot get rid on the 1, and the sin(x)^2, but I don't know how to get rid of them. Thank you.
• What you did is correct. You just need to learn or remember the trig identities to convert it.
tan²x + 1 = sec²x, cos²x + sin²x = 1, secx = 1/cosx

Let pick up from right before you got tan²x + 1 and do other approach by combining the fraction. Which they need to have the same denominator. We do that by multiplying cosx/cosx to the first term like you commented. You do not need to do the same to other terms if you are multiplying by 1 (but you can if you need to make the denominator the same). In this case, we don't need to.

= [cosx/cosx]+ [(sin²x)/(cos²x]
= [(cosx/cosx)(cosx/cosx)]+ [sin²x/cos²x]
= [cos²x/cos²x]+ [sin²x/cos²x]
= [cos²x + sin²x] / cos²x
= 1 / cos²x = sec²x = tan²x + 1

All 3 are equivalent and are acceptable, but if asked to simplified, it would probably be sec²x. It is good to memorize all 3 as the derivative of tanx instead of sec²x alone like in the textbook. All at least recognize they are equivalent.
• So around the mark, instead of using the trig identity, I cancelled out the cos^2(x) in the numerator with the cos^2(x) in the denominator to get an answer of 1 + sin^2(x). Is this wrong? And if so, why?
• I think you may have made a very common mistake in your algebra! You can only cancel something if your entire numerator is multiplied by it. I'll give you an example with some numbers. take (6+8)/2. If you evaluate the parentheses first, you get 14/2, which is just 7. If you manipulate it, though, you can get 2(3+4)/2, which when you cancel, gives you 3+4, which is also 7! However, you can't just cancel one of them, as in cancelling from (6+8)/2 to (3+8). That doesn't give you the right answer, you must divide both numbers in the numerator by the denominator.
• Why is cos+sin equal to 1?
• By the way, it's cos^2+sin^2=1. On the unit circle (x^2+y^2=1) each point on the circle can be represented by the point (cos(theta),sin(theta)) because sin(theta)=opposite/hypotenuse but the hypotenuse is the radius which is 1, and the opposite=y. Therefore, sin(theta)=y. You can do the same with cosine, but with x instead of y to get cos(theta)=x. Substitute those into the equation of the unit circle to get cos^2(theta)+sin^2(theta)=1.
• is there video for proving differentiation of sin and cos function ?
• Hello, I'm having difficulty with this section in Calculus. I am able to work out the derivatives fine but when I plug an x such as pi/6 into the equation on my calculator, the answer I get is nowhere near the provided choices.
(1 vote)
• Your calculator has a "degrees" mode and a "radians" mode. Make sure that your calculator is set to the appropriate mode setting for the number that you're working with.
• One of the derivative questions was d/dx(tan(x)) at 3pi/4. The answer was sec^2(x) and finally x = 2. My question was with fractions being 1/2/4 which I got as x = 1/8. What I did was took (1/2) / (4/1). Is there a fraction rule that says to take (1 / (1/2)) vs ((1/2) / 4)?
• If you do have the expression 1/2/4, you should evaluate from left-to-right. So this is (1/2)/4=1/8.

However, division by 4 shouldn't have showed up in your answer. You're correct that the derivative of tan(x) is sec²(x), or 1/cos²(x).

cos(3π/4)=-√2/2, so this equals 1/(-√2/2)²=1/(1/2)=2.
• could we also derive dcotx/dx by taking cotx = 1/tanx and applying the quotient rule on that?