Main content

## AP®︎ Calculus BC (2017 edition)

### Unit 3: Lesson 11

Trigonometric functions differentiation- Derivatives of tan(x) and cot(x)
- Derivatives of sec(x) and csc(x)
- Derivatives of tan(x), cot(x), sec(x), and csc(x)
- Worked example: Derivative of sec(3π/2-x) using the chain rule
- Differentiate trigonometric functions
- Differentiating trigonometric functions review

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

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

AP.CALC:

FUN‑3 (EU)

, FUN‑3.B (LO)

, FUN‑3.B.3 (EK)

Sal finds the derivatives of tan(x) and cot(x) by writing them as quotients of sin(x) and cos(x) and using quotient rule.

## Want to join the conversation?

- 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)?(0 votes)
- 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.(25 votes)

- 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.(6 votes)- 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.(2 votes)

- is there video for proving differentiation of sin and cos function ?(3 votes)
- There is a set of proofs on KA starting here:

https://www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/ab-derivtive-rules-opt-vids/v/derivative-of-sin-x

but, you need to be familiar with trig identities and this proof that the`lim x-> 0 [sin(x)/x] = 1`

:

https://www.khanacademy.org/math/calculus-home/limits-and-continuity-calc/squeeze-theorem-calc/v/sinx-over-x-as-x-approaches-0

There is another type of proof that depends on combining two different definitions of e^x and also uses imaginary numbers ... if you are interested, make a comment and I'll dig up links later this week (or post a proof if I can't find any good ones) ...(3 votes)

- So around the2:00mark, 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?(2 votes)
- 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.(4 votes)

- 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)?(1 vote)
- 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.(2 votes)

- Why is cos+sin equal to 1?(1 vote)
- 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.(1 vote)

- so is something like -sin^2 x the same as sin^-1 x?

is it the inverse or what?(1 vote) - On the AP Exam, would it be acceptable to just write d/dx[tan(x)]=sec^2(x) and similar for cotangent, or would we have to write out each step to reach those derivatives in order to receive the most points?(1 vote)
- I guess you have to do that. Also, if you know how to do it, just write it instead of taking risks.(1 vote)

- 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.(0 votes)
- 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.(3 votes)

- What about questions such as derivative of sec(x) with respect to tan(x) i.e. derivative based on trigonometric function? How to compute solutions for these?(0 votes)
- For trigonometric derivatives, there are two that you must memorize: d/dx sin(x)=cos(x) and d/dx cos(x)= -sin(x). Once you have these, you can get everything else using the product and quotient rules as needed.

For example, sec(x)=1/cos(x). As such, d/dx sec(x) = d/dx 1/cos(x). By the quotient rule, d/dx 1/cos(x) = [cos(x)**0 - 1**-sin(x)]/cos(x)^2 = sin(x)/cos(x)^2 = [sin(x)/cos(x)]*[1/cos(x)]=tan(x)*sec(x). The quotient rule videos will give a more general explanation.(1 vote)

## Video transcript

- [Voiceover] We already
know the derivatives of sine and cosine. We know that the derivative
with respect to x of sine of x is equal to cosine of x. We know that the derivative
with respect to x of cosine of x is equal
to negative sine of x. And so what we want to do in this video is find the derivatives of the
other basic trig functions. So, in particular, we
know, let's figure out what the derivative with respect to x, let's first do tangent of x. Tangent of x, well this is the same thing as trying to find the
derivative with respect to x of, well, tangent of x is just sine of x, sine of x over cosine of x. And since it can be expressed as the quotient of two functions, we can apply the quotient
rule here to evaluate this, or to figure out what this is going to be. The quotient rule tells us
that this is going to be the derivative of the top function, which we know is cosine of
x times the bottom function which is cosine of x, so times cosine of x minus, minus the top
function, which is sine of x, sine of x, times the derivative
of the bottom function. So the derivative of cosine
of x is negative sine of x, so I can put the sine of x there, but where the negative
can just cancel that out. And it's going to be over, over the bottom function squared. So cosine squared of x. Now, what is this? Well, what we have here, this
is just a cosine squared of x, this is just sine squared of x. And we know from the Pythagorean identity, and this is really just out of, comes out of the unit circle definition, the cosine squared of x
plus sine squared of x, well that's gonna be
equal to one for any x. So all of this is equal to one. And so we end up with one
over cosine squared x, which is the same thing as,
which is the same thing as, the secant of x squared. One over cosine of x is secant, so this is just secant of x squared. So that was pretty straightforward. Now, let's just do the inverse of the, or you could say the
reciprocal, I should say, of the tangent function,
which is the cotangent. Oh, that was fun, so let's do that, d dx of cotangent, not cosine, of cotangent of x. Well, same idea, that's the
derivative with respect to x, and this time, let me make some
sufficiently large brackets. So now this is cosine of x over sine of x, over sine of x. But once again, we can use
the quotient rule here, so this is going to be the
derivative of the top function which is negative, use that magenta color. That is negative sine of x times the bottom function, so times sine of x, sine of x, minus, minus the top function, cosine of x, cosine of x, times the
derivative of the bottom function which is just going to
be another cosine of x, and then all of that over
the bottom function squared. So sine of x squared. Now what does this simplify to? Up here, let's see, this
is sine squared of x, we have a negative there, minus cosine squared of x. But we could factor out the negative and this would be
negative sine squared of x plus cosine squared of x. Well, this is just one by
the Pythagorean identity, and so this is negative
one over sine squared x, negative one over sine squared x. And that is the same thing as negative cosecant squared, I'm running out of space, of x. There you go.