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## AP®︎ Calculus BC (2017 edition)

### Unit 4: Lesson 5

Derivatives of inverse trigonometric functions

# Differentiating inverse trig functions review

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.E (LO)
,
FUN‑3.E.2 (EK)
Review the derivatives of the inverse trigonometric functions: arcsin(x), arccos(x), and arctan(x).

## What are the derivatives of the inverse trigonometric functions?

start fraction, d, divided by, d, x, end fraction, \arcsin, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, square root of, 1, minus, x, squared, end square root, end fraction
start fraction, d, divided by, d, x, end fraction, \arccos, left parenthesis, x, right parenthesis, equals, minus, start fraction, 1, divided by, square root of, 1, minus, x, squared, end square root, end fraction
start fraction, d, divided by, d, x, end fraction, \arctan, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, 1, plus, x, squared, end fraction

## Want to join the conversation?

• Why were arccsc(x) & arcsec(x) left out? Often ignored, but they're apart of the trig family too!
• for the ones left out.
y = arcsec(x)
sec y = x
take the derivative of both sides
(sec y)(tan y)(dy/dx) = 1
dy/dx = 1/((sec y)(tan y))
We can replace sec y with x
dy/dx = 1/(x*tan y)
now simply use the Pythagorean identity
1 + tan^2(y) = sec^2(y)
solve for tan(y)
tan(y)= sqrt(sec^2(y)-1)
so
dy/dx = 1/(x*tan y) now becomes
dy/dx = 1/(x*sqrt(sec^2(y)-1))
and finally
dy/dx = 1/(x*sqrt(x^2-1))

For arccsc(x) you do the same thing and make use of the identity
cot^2(x) + 1 = csc^2(x)

For cot(x), make use of this same identity. Ultimately all 6 inverse trig functions can be found with this same methodology using the 3 different pythagorean identities. The method used for arctan(x) would have been easier if he simply divided both sides by sec^2(x) and then replaced sec^2(x) with 1+tan^2(x)
• What about the reciprocal trig inverses? Can you provide videos/text about arccot(x), for example?
• d/dx arccot(x) = - 1 / (1+x²)

d/dx arcsec(x) = 1 / (x√(x²-1)) ; for 0≤x<π/2 and π≤x<3π/2
d/dx arcsec(x) = 1 / (|x|√(x²-1)) ; for 0≤x<π/2 and π/2<x≤π

d/dx arccsc(x) = - 1 / (x√(x²-1)) ; for 0≤x<π/2 and π≤x<3π/2
d/dx arccsc(x) = - 1 / (|x|√(x²-1)) ; for 0≤x<π/2 and π≤x<3π/2

We often use the first case in college however. The quadrants determine tan function positive or negative in the differentiation. The first restriction is QI and QIII, so tan is always positive, thus we have x without the absolute value before the radical. The second restriction is QI and QII, tan can either be positive or negative, thus we have |x|.

Another thing to remember that the derivatives of the "co-" arc-trig functions is just the negative of their counterparts. See how the derivative of arccos(x) is just negative of what arcsin(x) has, similar for arctan(x) and arccot(x), and arcsec(x) and arccsc(x)
• Why do we call inverse trig functions as arctrig functions?
• You see regular trig functions represent a ratio. Arctrig functions represent an angle. In a way, an arc is an angle which has been given an extra dimension of radius. If you were to zoom out while looking at an arc it will look like an angle.
• could you give an example on how to solve more difficult questions? for example find the derivative of : arcsin(x) / arcsin (2x).
• how can this be applied in real life?
• Maybe someones heartbeat can be represented by a trigonometric function, and you want need to report to a doctor the rate at which the patient's heart rate is increasing at a moment in time, so the doctor can perform his procedure when the rate is calm and steady. (I don't know the exact job of a doctor, but something like that)
• I can get all the answers correct now. But How long will it take before I forget all about it?
• I am assuming that you are asking about remembering formulas for differentiating inverse trig functions.

If you forget one or more of these formulas, you can recover them by using implicit differentiation on the corresponding trig functions.

Example: suppose you forget the derivative of arctan(x). Then you could do the following:

y = arctan(x)
x = tan(y)
1 = sec^2(y) * dy/dx
dy/dx = 1/sec^2(y)
dy/dx = 1/[tan^2(y) + 1]
dy/dx = 1/(x^2 + 1).

So the derivative of arctan(x) is 1/(x^2 + 1).
• What about the derivative of arccot(x)?
(1 vote)
• let y = arccot(x), then cot(y) = x, and d/dx(cot(y)) = 1, csc^2(y) * dy/dx = 1, dy/dx = sin^2(y).
plug in x = arccot(x), we have dy/dx = [sin(arccot(x))]^2
let opp = 1, and adj = x, sin(angle) = 1/sqrt(x^2+1)
therefore sin(arccot(x)) = 1/sqrt(x^2+1).
Hence, dy/dx = [1/sqrt(x^2+1)]^2 = 1/(x^2+1)
• I am wondering, if not going straight for the rules above, how I can figure out the derivative of arcsin(-3x). I was hoping to use the trig rule and chain rule but then I got stuck...
(1 vote)
• f(x) = arcsin(u) and u = -3x
Using the arcsin trig rule and chain rule:
f'(x) = d/dx (arcsin(-3x)) * du/dx = (1/√(1-(-3x)²)) * -3 = -3/√(1-9x²)
• Through a calculator, I came up with some "master formulas", if you will. Are these correct?

d/dx(arcsin Ax/B)= A/(B√(1-(A^2*x^2)/B^2))
d/dx(arccos Ax/B)= -A/(B√(1-(A^2*x^2)/B^2))
d/dx(arctan Ax/B)= A/(((A^2*x^2)/B^2)+B)
(1 vote)
• Showing your reasoning always help. Based on the functions evaluated you have a rational function inside an inverse trigonometric. You just have to go to the formulas for the inverse trigonometric and replace the u values with your rational function. Then apply chain rule and see wether they are correct or not. You can also input them on WA (Wolfram Alpha). I'm going to do the first one for you.

We know that (arcsin u)' is 1/√(1-u^2) · u'
Not going to show the informal nor formal proof.
I'm assuming a and b are coefficients ∊ ℝ.
(arcsin ax/b)' = 1/√(1-[ax/b]^2) · (ab - 0)/b^2
= a/b√(1-[ax/b]^2)

So seems like what you have done in the three cases is wrong. Don't take my word, show your reasoning and it will be more clear for you in the case you made an error along the way.