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# Maths Help! watch

1. Thank you.
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2. 1. y = arcsec x
x = sec y
dx/dy = sec y tan y
dy/dx = 1/(sec y tan y) = 1/(x tan y)
Since (tan y)^2 = (sec y)^2 - 1 and y > 0, tan y = ((sec y)^2 - 1)^1/2) = (x^2 - 1)^1/2
So dy/dx = 1/[x(x^2 - 1)^1/2]

2. Use a similar method, start with y = arcot x, x = cot y then go from there
3. hmmm.. this is probably P5.. right? I havent done this but I thought I'd have a go..

y=sec^-1 x

sec y = x

sec y tan y dy/dx = 1

dy/dx = 1/(sec y tan y)

now here it goes a bit wobbly..
y=sec^-1 x

dy/dx = 1/(sec(sec^-1 x) tan(sec^1))

.. which doesnt look good
if you go to http://www.calc101.com/webMathematic...es.jsp#topdoit and paste in sec^-1[x] you get the answer, which isn't 100 miles off.. if there's someone cleverer here they can probably see what i'm doing wrong
4. Kimoni - the first part of your working is right, because you have a sec y at the end this can be replaced with x, for the tan y you need to use the trig identity 1 + (tan y)^2 = (sec y)^2 I think that in every arc trig differentiation you need to use an identity at some point. Conveniently though the restricted domain always means you don't need to worry about plus or minus when you square root because you'll only need the positive square root.
5. (Original post by Bezza)
Kimoni - the first part of your working is right, because you have a sec y at the end this can be replaced with x, for the tan y you need to use the trig identity 1 + (tan y)^2 = (sec y)^2 I think that in every arc trig differentiation you need to use an identity at some point. Conveniently though the restricted domain always means you don't need to worry about plus or minus when you square root because you'll only need the positive square root.
aah.. once I read your answer I got it.. It's just getting your brain to recognise things like sec y can be substituted for x.. yup I see it now, did the rest on my own ta!
6. d(cot^-1 x)/dx

let y=cot^-1 x
cot y = x

differentiate:
-(cosec^2 y)*dy/dx = 1

dy/dx=-1/(cosec^2 y)

dy/dx= -sin^2 y

because tan x = sin x /cos x
tan x cos x = sin x

so -sin^2 x =-(tan^2 x)*(cos^2 x)

let (cos^2 x) = 1/sec^2 x = 1/(tan^2 x +1)

so dy/dx=-tan^2x/(tan^2x + 1)

going back to x = cot y
let x = 1/tan y
so tan y = 1/x
therefore tan^2 y = x^-2

dy/dx = -x^-2 / x^-2 + 1
simplify to: -1/(x^2 + 1)

which I've checked at calc101

phew. I betcha there's a simpler way of doing that than what I just did
7. (Original post by kimoni)
d(cot^-1 x)/dx

let y=cot^-1 x
cot y = x

differentiate:
-(cosec^2 y)*dy/dx = 1

dy/dx=-1/(cosec^2 y)
You're right, there is a simpler way! (cosec y)^2 = 1 + (cot y)^2 = 1 + x^2 so dy/dx = -1/(1+x^2)

In questions like this you normally use the identity (sin y)^2 + (cos y)^2 = 1, or the others derived from this by dividing through by (sin y)^2 or (cos y)^2

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