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Differentiating inverse trig functions

Find the maximum value of f(x) = (sin^-1x)^2*(cos^-1x) in the interval 0 <x < 1

So I used product rule and got:

f'(x)=[2sin^-1xcos^-1x-(sin^-1x)^2]/ (1-x^2)^1/2.

Then I was unsure about the maximum value part I tried letting the numerator = 0 but didn't get far. Answer is pi^3/54

(edited 8 years ago)

Reply 1

TeeEm

Original post by Super199

very nice question.

Stolen already.

Reply 2

Student403

Original post by TeeEm

very nice question.

Stolen already.

Do you know what module this is taught in?

Reply 3

TeeEm

Original post by Student403

Do you know what module this is taught in?

this has to be Further maths these days

FP2 in AQA, MEI, OCR

FP3 in EDEXCEL, CCEA

cannot quite remember WJEC

Reply 4

Student403

Original post by TeeEm

this has to be Further maths these days

FP2 in AQA, MEI, OCR

FP3 in EDEXCEL, CCEA

cannot quite remember WJEC

I'm Edexcel so this'll be FP3 for me. Thanks! Wonder how hard it is

Reply 5

TeeEm

Original post by Student403

I'm Edexcel so this'll be FP3 for me. Thanks! Wonder how hard it is

This is a beautiful question but not to the style of EDEXCEL

Reply 6

Student403

Original post by TeeEm

This is a beautiful question but not to the style of EDEXCEL

Fair enough - I'll have to see for myself!

Reply 7

Craig1998

Original post by Student403

I'm Edexcel so this'll be FP3 for me. Thanks! Wonder how hard it is

I covered this this week (FP3 Edexcel) and it isn't as horrid as this question sets it out to be.

As a suggestion to the answer, the denominator obviously vanishes which means the numerator is to be set = to 0. Could you collect like terms?

Spoiler

The bit on the right looks tricky, but left can give you a solution.

Edit, oh no it cant. Damn limits.

Reply 8

Super199

Original post by Craig1998

Spoiler

The bit on the right looks tricky, but left can give you a solution.

Edit, oh no it cant. Damn limits.

Yeah I factorised the top bit but then obviously couldn't get anywhere with it lol

Reply 9

TeeEm

Original post by Craig1998

Spoiler

The bit on the right looks tricky, but left can give you a solution.

Edit, oh no it cant. Damn limits.

this question is a little bit more than differentiation.

I would love to see how it is phrased

Reply 10

davros

Original post by Super199

Original post by TeeEm

very nice question.

.

This is indeed a very nice question.

And no need to differentiate any inverse trig functions at all :smile:

Reply 11

Super199

Original post by davros

This is indeed a very nice question.

And no need to differentiate any inverse trig functions at all :smile:

lol what. Explain please :smile:

Reply 12

davros

Original post by Super199

lol what. Explain please :smile:

You want me to give the question away and spoil your enjoyment? :biggrin:

For 0 < x < 1 there is a very simple relationship between

sin^{-1}x

and

\cos^{-1}x

. Do you know what it is? If not, draw a right angled triangle with hypotenuse 1 and one other side labelled x, and work out what those two functions are in terms of angles in the triangle.

If you now write

u = sin^{-1}x

your original function can be written as a new function g(u) solely in terms of u. You can now reduce the problem to finding the max of g(u) as u varies. This is a C1 task :biggrin:

I'll leave you to fill in the details :smile: