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    how do you show that:

    cos(sin^-1 x) = ±√(1-x^2), where ^-1 -> inverse
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    let u = sin-1 x
    x = sinu
    cos u = rt(1 - sin²u)
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    or draw a right-angled triangle with opposite length x and hypotenuse 1
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    (Original post by Fermat)
    let u = sin-1 x
    x = sinu
    cos u = rt(1 - sin²u)
    how dumb of me! i was so depressed about that that i couldn't sleep! aghhh...i tried integrating and all this werid crap, lol

    cheers:cheers:

    pk
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    need help with another maths quesiton form problem sheet 2:

    show that:

    dn(xn)=n!
    dxn

    any ideas?

    cheers

    PK
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    (Original post by Phil23)
    need help with another maths quesiton form problem sheet 2:

    show that:

    dn(xn)=n!
    dxn

    any ideas?

    cheers

    PK
    Let y = xn:

    dy/dx = nxn-1
    d2y/dx2 = n(n-1)xn-2
    d3y/dx3 = n(n-1)(n-2)x(n-3)
    ...

    The nth derivative will leave x0, because each time the power will be reduced by one. So the result will be [n(n-1)(n-2)...(3)(2)(1)] x x0 = n!


    It's not really a proof as such, but it shows how the result falls out.
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    (Original post by Phil23)
    need help with another maths quesiton form problem sheet 2:

    show that:

    dn(xn)=n!
    dxn

    any ideas?

    cheers

    PK
    You could use induction to find the kth derivative of x^n and then set k=n, showing that the term is n!
    You could guess and prove that the kth derivative of x^n is \frac{n!}{(n-k)!}x^{n-k}
    Obviously then setting k=n gives \frac{n!}{0!}x^{0}=n!
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    (Original post by Gaz031)
    You could use induction to find the kth derivative of x^n and then set k=n, showing that the term is n!
    You could guess and prove that the kth derivative of x^n is \frac{n!}{(n-k)!}x^{n-k}
    Obviously then setting k=n gives \frac{n!}{0!}x^{0}=n!
    #the prob is that the function varies with the derivative...according to the sheet..i.e. 10th derivative of x^10, and 9th of x^9 etc...so a bit lost...tried doing a generalised chain rule, but not got anywhere yet

    pk

    must be easier than induction
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    (Original post by Phil23)
    #the prob is that the function varies with the derivative...according to the sheet..i.e. 10th derivative of x^10, and 9th of x^9 etc...so a bit lost...tried doing a generalised chain rule, but not got anywhere yet

    pk

    must be easier than induction
    How does that matter? If you let n=k then you obviously have the nth derivative of x^n which can mean the 10th derivative of x^10, the 9th derivative of x^9... and so on for any positive integer n.
    Of course, you could just 'explain why' but perhaps this way is more likely to be regarded as a proper proof.
 
 
 
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