Q: 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)

Q: could you give an example on how to solve more difficult questions? for example find the derivative of : arcsin(x) / arcsin (2x).

In order to do this you need to know 1) The derivative of arcsin(x) - covered in this video. 2) The chain rule (to get the arcsin(2x) bit)- covered here: https://www.khanacademy.org/math/ap-calculus-ab/product-quotient-chain-rules-ab/chain-rule-ab/v/chain-rule-introduction And 3) the quotient rule - covered here: https://www.khanacademy.org/math/ap-calculus-ab/product-quotient-chain-rules-ab/quotient-rule-ab/v/quotient-rule-from-product-rule

Q: 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.

Q: 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...

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)

Q: 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).

Q: please prove the case when x>0 , y<0 and xy<-1 then: arctan(x) - arctan(y) = pi + arctan[(x-y)/(1+xy)]

```arctan(x) - arctan(y) = pi + arctan[(x-y)/(1+xy)]``` ```tan(arctan(x) - arctan(y)) = tan(pi + arctan[(x-y)/(1+xy)])``` ```(tan(arctan(x)) - tan(arctan(y)))/(1 + tan(arctan(x)) * tan(arctan(y))) = (tan(pi) + tan(arctan[(x-y)/(1+xy)]))/(1 + tan(pi) * tan(arctan[(x-y)/(1+xy)]))``` ```(x - y)/(1 + xy) = (0 + [(x - y)/(1 + xy)])\(1 + (0)[(x + y)/(1 + xy)])``` ```(x - y)/(1 + xy) = (x - y)/(1 + xy)``` You don't need to prove it for individual cases, as it is true for all x and y, xy != -1.

Q: what is the derivertive of y=tan inverse 5x+1/2?

you can use the chain rule, so set f(x)=arctan(x) and set g(x)=5x+1/2. Then use the chain rule and you will get dy/dx= 5/1+((5x+1/2)^2)

Q: can any one help with this y= arctan(a.x+b)

𝑑∕𝑑𝑥(arctan 𝑥) = 1∕(1 + 𝑥²) Using the chain rule, we then find 𝑑∕𝑑𝑥(arctan(𝑎𝑥 + 𝑏)) = 1∕(1 + (𝑎𝑥 + 𝑏)²)⋅𝑑∕𝑑𝑥(𝑎𝑥 + 𝑏) = 𝑎∕(1 + (𝑎𝑥 + 𝑏)²)

Question 1

Why were arccsc(x) & arcsec(x) left out? Often ignored, but they're apart of the trig family too!

Accepted Answer

Probably because it's actually really confusing. Think about it: Take arcsec(x). d/dx (1/cos(x)) would be a quotient of derivatives. I presume you know the complicated equation for that. Stuff arcsec(x) into it. Yeah. Also you'd probably rarely see it on the AP test.

Question 2

What about the reciprocal trig inverses? Can you provide videos/text about arccot(x), for example?

Accepted Answer

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)

Question 3

how can this be applied in real life?

Accepted Answer

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)

Question 4

could you give an example on how to solve more difficult questions? for example find the derivative of :    arcsin(x) / arcsin (2x).

Accepted Answer

In order to do this you need to know
1) The derivative of arcsin(x) - covered in this video.

2) The chain rule (to get the arcsin(2x) bit)- covered here: https://www.khanacademy.org/math/ap-calculus-ab/product-quotient-chain-rules-ab/chain-rule-ab/v/chain-rule-introduction

And 3) the quotient rule - covered here: https://www.khanacademy.org/math/ap-calculus-ab/product-quotient-chain-rules-ab/quotient-rule-ab/v/quotient-rule-from-product-rule

Question 5

Why do we call inverse trig functions as arctrig functions?

Accepted Answer

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.

Question 6

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...

Accepted Answer

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)

Question 7

I can get all the answers correct now. But How long will it take before I forget all about it?

Accepted Answer

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).

Question 8

please prove the case when x>0 , y<0 and xy<-1

then:  arctan(x) - arctan(y) = pi + arctan[(x-y)/(1+xy)]

Accepted Answer

```arctan(x) - arctan(y) = pi + arctan[(x-y)/(1+xy)]```
```tan(arctan(x) - arctan(y)) = tan(pi + arctan[(x-y)/(1+xy)])```
```(tan(arctan(x)) - tan(arctan(y)))/(1 + tan(arctan(x)) * tan(arctan(y))) = (tan(pi) + tan(arctan[(x-y)/(1+xy)]))/(1 + tan(pi) * tan(arctan[(x-y)/(1+xy)]))```
```(x - y)/(1 + xy) = (0 + [(x - y)/(1 + xy)])\(1 + (0)[(x + y)/(1 + xy)])```
```(x - y)/(1 + xy) = (x - y)/(1 + xy)```
You don't need to prove it for individual cases, as it is true for all x and y, xy != -1.

Question 9

what is the derivertive of y=tan inverse 5x+1/2?

Accepted Answer

you can use the chain rule, so set f(x)=arctan(x) and set g(x)=5x+1/2. Then use the chain rule and you will get dy/dx= 5/1+((5x+1/2)^2)

Question 10

can any one help with this
y= arctan(a.x+b)

Accepted Answer

𝑑∕𝑑𝑥(arctan 𝑥) = 1∕(1 + 𝑥²)

Using the chain rule, we then find
𝑑∕𝑑𝑥(arctan(𝑎𝑥 + 𝑏))
= 1∕(1 + (𝑎𝑥 + 𝑏)²)⋅𝑑∕𝑑𝑥(𝑎𝑥 + 𝑏)
= 𝑎∕(1 + (𝑎𝑥 + 𝑏)²)

Differential Calculus

Course: Differential Calculus > Unit 3

Differentiating inverse trig functions review

What are the derivatives of the inverse trigonometric functions?

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