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Maths Question

27FFBF79-2839-4D22-AA6D-3AC79CDB4580.jpg.jpeg

Anyone know how to do these?

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Reply 1
Avatar for mqb2766
mqb2766
21
You're given the function, so calculate the dy/dx and d^2y/dx^2 and show they satisfy the stated identity.
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
would it be sin^2x = y

Then d/dx both sides?
Reply 4
Avatar for mqb2766
mqb2766
21
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?
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?
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
Original post by PARAMLUHADIYA1


then differentiate to get 1/2sqrt(y) * y' = 1/sqrt(1-x^2)



How did you get 1/sqrt(1-x^2) ?
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)
Reply 9
Avatar for davros
davros
16
Original post by PARAMLUHADIYA1
I'm going to be using y' = dy/dx and y'' = d2y/dx2


Original post by PARAMLUHADIYA1
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.
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
Original post by PARAMLUHADIYA1
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
Original post by PARAMLUHADIYA1

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’ ?
(edited 3 years ago)
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
Original post by PARAMLUHADIYA1
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?
Reply 15
Avatar for davros
davros
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 :smile:
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 :smile:


Do you know how to do 38?
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?
Reply 18
Avatar for davros
davros
16
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 :smile:

Post your working if you're stuck.
BD6A3BF8-9642-4C3C-BCD9-724E06DEBD9C.jpg.jpeg so confused?

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