# Maths Question

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#1

Anyone know how to do these?
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1 month ago
#2
You're given the function, so calculate the dy/dx and d^2y/dx^2 and show they satisfy the stated identity.
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#3
(Original post by mqb2766)
You're given the function, so calculate the dy/dx and d^2y/dx^2 and show they satisfy the stated identity.
How would I go about calculating dy/dx of (arcsinx)^2
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#4
would it be sin^2x = y

Then d/dx both sides?
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1 month ago
#5
(Original post by CaptainDuckie)
would it be sin^2x = y

Then d/dx both sides?
You seem to have just replaced asin with sin?
You could root the original function then take sin?
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#6
(Original post by mqb2766)
You seem to have just replaced asin with sin?
You could root the original function then take sin?
so root y = sin?
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1 month ago
#7
I'm going to be using y' = dy/dx and y'' = d2y/dx2
For the first one, root both sides to get sqrt(y) = arcsin(x)
then differentiate to get 1/2sqrt(y) * y' = 1/sqrt(1-x^2)
now square both sides and rearrange to get y'^2(1-x^2) = 4y
then differentiate again to get 2y'y''(1-x^2) + y'^2(-2x) = 4y'
then remove the common factors (y' and 2) to get y''(1-x^2) -y'x -2 = 0
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#8

then differentiate to get 1/2sqrt(y) * y' = 1/sqrt(1-x^2)
How did you get 1/sqrt(1-x^2) ?
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1 month ago
#9
(Original post by CaptainDuckie)
How did you get 1/sqrt(1-x^2) ?
d/dx (arcsin(x)) = 1/sqrt(1-x^2) is a standard result I think
The proof goes like this:
y = arcsin(x)
sin(y) = x
cos(y)dy/dx = 1
dy/dx = 1/cos(y)
and cos(y) = sqrt(1-x^2) since x = sin(y)
so dy/dx = 1/sqrt(1-x^2)
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1 month ago
#10
I'm going to be using y' = dy/dx and y'' = d2y/dx2
d/dx (arcsin(x)) = 1/sqrt(1-x^2) is a standard result I think
The proof goes like this:
Please don't post full solutions - it's against the rules of the forum!!

OP - do you know the derivative of arcsin x? As stated above, it's a standard result, but is easy enough to derive if you don't know it.
You will then need to apply the chain rule to the original function.
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1 month ago
#11
(Original post by davros)
Please don't post full solutions - it's against the rules of the forum!!
Oh sorry, I'm kinda new here, I'll keep that in mind
1
#12
d/dx (arcsin(x)) = 1/sqrt(1-x^2) is a standard result I think
The proof goes like this:
y = arcsin(x)
sin(y) = x
cos(y)dy/dx = 1
dy/dx = 1/cos(y)
and cos(y) = sqrt(1-x^2) since x = sin(y)
so dy/dx = 1/sqrt(1-x^2)
Complete legend, thank you so much
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#13
then differentiate to get 1/2sqrt(y) * y' = 1/sqrt(1-x^2)
Wouldn’t the sqrt y different into 1/2 y^ -1/2 * y’ ?
Last edited by CaptainDuckie; 1 month ago
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1 month ago
#14
(Original post by CaptainDuckie)
Wouldn’t the sqrt y different into 1/2 y^ -1/2 * y’ ?
Yup that's what I meant. Typing things out here is a pain lmao I think next time I'll just write on paper and upload an image
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#15
Yup that's what I meant. Typing things out here is a pain lmao I think next time I'll just write on paper and upload an image
Can you write it out on paper please?
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1 month ago
#16
(Original post by CaptainDuckie)
Can you write it out on paper please?
Please note that we do NOT give full solutions on TSR - it is against the rules!

We can provide you with hints when you are stuck, but you should be doing the work 0
#17
(Original post by davros)
Please note that we do NOT give full solutions on TSR - it is against the rules!

We can provide you with hints when you are stuck, but you should be doing the work Do you know how to do 38?
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1 month ago
#18
(Original post by CaptainDuckie)
Do you know how to do 38?
Similar way, using the derivative of arctan.
Why not upload your attempt, based on what you've learnt from the previous question?
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1 month ago
#19
(Original post by CaptainDuckie)
Do you know how to do 38?
38 is actually quite straightforward - as long as you know the (standard) derivative of arctan x, the second derivative is easy and the algebra drops out in a few lines Post your working if you're stuck.
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#20
so confused?
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