Question

Calc 2 Help: Derivative of Inverse Trig Functions
y = xarcsin(x/4) + sqrt(16-x^2)

Answer 1

Dear Agent-N:
Appreciate the way you use parentheses appropriately to ensure proper interpretation of these math expressions.
It appears to me that your xarcsin(x/4) is a product. Correct?

Answer 2

Yes! That is correct.

Answer 3

Do you have available a reference work that lists the derivatives of inverse trig functions?

Answer 4

I know the derivatives of arcsin, arccos, and arctan.

Answer 5

The derivative with respect to x of arcsin(x) is 1 over Sqrt(1-x^2). Agreed?

Answer 6

don't we need the product rule as well for xarcsin(x/4)?

Answer 7

like f g' + g f'?

Answer 8

Oh. I forgot to include the rest of the question. That is the function, but what they want me to do is find y'(2)

Answer 9

o-o

Answer 10

oh I see we need the derivative first and then plug in y'(2)

Answer 11

Yes, of course; but the focus of this problem is "inverse trig functions," and so I'm reviewing the derivative of arcsin x before applying the product rule.

Answer 12

I used the product rule, but I still have a arcsin in the derivative that I don't know how to handle.

Answer 13

it should be in your textbook. they're given in a chart.

Answer 14

http://upload.wikimedia.org/math/9/a/6/9a6ca82be93f7fd7a2b8d102a308b547.png
use the first one

Answer 15

That makes complete sense, AgentN, that you still have the arcsin function in your expression. Simply leave it as is there; you are writing the derivative of the arcsin function in another term, right?

Answer 16

I know the derivatives of inverse trig functions. That is not the problem.

Answer 17

Would you mind defining what IS posing a problem for you?

Answer 18

Not in another term. I need to calculate the value of the expression at 2. After finding the derivative, of course.

Answer 19

I guess the arcsin is left alone.. like we leave the x alone deal with the arcsin + leave the arcsin alone deal with the x

Answer 20

and then plug in ... oh I see how to calculate arcsin

Answer 21

Here. Let me tell you what I have right now.

Answer 22

calculating the arcsin part when you put y'(2)....

Answer 23

Good approach. It'd help me tremendously to see what you have done so far.

Answer 24

x/[4sqrt(1-x^2)] + arcsin(x/4) - 1/sqrt(16-x^2)

Answer 25

That is the derivative I came up with, but I don't know how to plug in 2 to arcsin.

Answer 26

2/[4sqrt(1-2^2)] + arcsin(2/4) - 1/sqrt(16-2^2)

Answer 27

2/[4sqrt(1-4)] + arcsin(1/2) - 1/sqrt(12)

Answer 28

arcsin(2/4) is fine... arcsin2 is not.

Answer 29

I believe you are completely on the right track there. Now you just want to evaluate this expression at x=2, right?

Answer 30

Oh, wait.

Answer 31

2/[4sqrt(-3)] + arcsin(1/2) - 1/sqrt(12)

Answer 32

Yes. And my other problem is the negative sign under the radical.

Answer 33

Yeah, arcsin(1/2) is fine.... the square root, this just means the derivative doesn't exist at x=2.

Answer 34

derivative doesn't exist at x = 2... because of that negative sign?

Answer 35

Assuming the derivative is calculated correctly...

Answer 36

Let's focus on the very first term.

x/[4sqrt(1-x^2)] + arcsin(x/4) - 1/sqrt(16-x^2)

Replacing x with 2 results in a first term that looks like

2
----------------------

2*Sqrt(1-(2/4)^2)

Think: where did that (2/4) come from, Agent N? Yes, assuming that the derivative was calculated correctly, agentOsmith..

Answer 37

Agent N: Please quickly review your original statement of this homework problem and note that the argument of arcsin is (x/4), not just x.

Answer 38

^i don't think it's correct.

derivative of arcsin(x/4) = 1/sqrt ( 1-(x/4)^2 )

Answer 39

Definitely x/4!

Answer 40

Thus, if x=2, we end up with arcsin (2/4) = arcsin (1/2) = pi/6 radians.

Answer 41

Are you sure, agent0smith?? I thought about that, but inserting x/4 as the input to the inverse function would make it a chain rule. Right?

Answer 42

uhh isn't sin 30 = pi/6
that's different than arcsin

Answer 43

Oh yeah i forgot about that.

1/4*1/sqrt ( 1-(x/4)^2 )

Answer 44

=1/sqrt(16 - x^2)

Answer 45

Once we have a tentative solution, I'd suggest that you, Agent N, quickly re-do the problem to ensure that all the differentiation (including application of the chain rule) has been done correctly.

Answer 46

Check it here: http://www.wolframalpha.com/input/?i=arcsin%28x%2F4%29++derivative

Answer 47

Do you think we'll figure it out within 20 minutes? Because that's when this question is due :(

Answer 48

:O

Answer 49

^yes. we have already.

Answer 50

arcsin (1/20 = Pi/6 (an angle)
sin (Pi/6) = sin (30 deg) = 1/2 (a ratio).

Answer 51

I see agentsmith. I completely forgot about good ole wolfphram.

Answer 52

So what is the final answer then?

Answer 53

OK, Agent N: seeing that you have 3 helpers at once, perhaps it'd be good for you to sit back for a moment and tell us exactly what it is that we can help you to finish this problem solution.

Answer 54

Is my derivative correct? WIth the negative under the radical symbol?

Answer 55

Derivative of:
y = xarcsin(x/4) + sqrt(16-x^2)

y' = arcsin(x/4) +x*1/sqrt(16 - x^2) + 0.5*2x/sqrt(16-x^2) w/o really

simplifying. Then plug in x=2

Answer 56

And the 2/4 as the arcsin input?

Answer 57

There is no negative sign

Answer 58

The 2/4 as input to the arcsin function is just fine.

I think you're referring to the derivative of arcsin (x-4), which DOES have a negative sign under the radical.

Answer 59

or alternatively: http://www.wolframalpha.com/input/?i=Derivative+of%3A+y+%3D+xarcsin%28x%2F4%29+%2B+sqrt%2816-x%5E2%29
now plug in x=2

Answer 60

I see. I had the wrong derivative of arcsin(x/4). That takes care of it. I think. Let me calculate it fully now.

Answer 61

Excuse me: you've probably noticed I've made a typo or two. arcsin (x/4) is correct.

Answer 62

Yes, why don't you take your time and then ask questions about anything remaining unclear to you.

Answer 63

Yes. I'll be back in just one second.

Answer 64

Shoot. I'm not getting it.

Answer 65

I wish the draw function would come back T_T... can you scan what you have and attach it on here?

Answer 66

As before, please pinpoint where you need help.

Answer 67

I plugged in 2 for the wolfphram alpha derivative and I got a string of radical expressions and pi/6 that wasn't correct :(

Answer 68

My arithmetic might be wrong.

Answer 69

What can I/we do to help you? Please take charge.

Answer 70

They want y'(2), correct?

Answer 71

k k kk k k the best thing to do right now. is to take the derivative.. product rule slowly.... for the first one.

Answer 72

Yes: first find the derivative of the given function, and then evaluate that derivative at x=2.

Answer 73

Yes. What is the final value.

Answer 74

I have 5 minutes left.

Answer 75

:S the pressure... ok ok .. errr

Answer 76

how many tries?

Answer 77

Im usually all for taking my time, but I spent an hour on this.. I will make sure to understand it later but I need help

Answer 78

arcsin(1/2) = ?
if theta = arcsin(1/2)
then sin theta = 1/2
so theta = 30 degrees or pi/6

Answer 79

inkyvoyd! 3 tries left.

Answer 80

I think it's one of those timed assignments like mymathlab

Answer 81

It's webassign. The due date for the assignment is midnight.

Answer 82

I think I'll have to give this one up.

Answer 83

I just wrote the solution!

Answer 84

screenshot and crop the problem?

Answer 85

The final answer is not pi/6. What about the rest of it?

Answer 86

I KNOW! @dan815 SAVE US D:

Answer 87

What rest? It is pi/6.

Answer 88

http://screencast.com/t/NPK40sA4ltE4

Answer 89

and you put down just pi/6 and it was wrong?

Answer 90

THE FINAL ANSWER IS pi/6 AND NOTHING ELSE

Answer 91

Oh....you were right agent smith. I'm sorry...

Answer 92

lol i gave you this:

or alternatively: http://www.wolframalpha.com/input/?i=Derivative+of%3A+y+%3D+xarcsin%28x%2F4%29+%2B+sqrt%2816-x%5E2%29
now plug in x=2

Answer 93

THANK YOU GUYS SO MUCH. I'm still not sure WHY it's pi/6, but I have other obligations for the evening, so I will look into it tomorrow morning. Thanks again for all your help!!

Answer 94

y=xarcsin(x/4) + sqrt(16-x^2)

I see why. We forgot a negative.

y' = arcsin(x/4) + 1/4*x/sqrt(1-(x/4)^2) + 0.5*(-2x)/sqrt(16-x^2) (those last two terms are the same but diff signs, so they cancel)...
y' = arcsin(x/4)

Answer 95

Thank you very much for your extra effort, AOS.

Answer 96

^i only noticed when checking the solution on W.A for how they simplified it... i wondered why they had a negative sign. Derivative of -x^2 is -2x... jeez.