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    How would I do d2y / dx2 fpr y = (sin-1x)2?

    for dy/dx I got 2sin-1x / (1-x2)1/2

    and for d2y / dx2 I got 2-xsin-1x / (1-x2) 3/2

    the answer is supposed to be 2( (1-x2)1/2 + xsin-1x) / (1-x2) 3/2

    could someone explain how to get the correct answer

    thanks
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    (Original post by bl64)
    How would I do d2y / dx2 fpr y = (sin-1x)2?

    for dy/dx I got 2sin-1x / (1-x2)1/2

    and for d2y / dx2 I got 2-xsin-1x / (1-x2) 3/2

    the answer is supposed to be 2( (1-x2)1/2 + xsin-1x) / (1-x2) 3/2

    could someone explain how to get the correct answer

    thanks
    please post a picture of your workings
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    I think you forgot that you need to use the product rule for the second derivative in this case. The 2x*arcsinx(1-x^2)^-3/2 is a product meaning that you cannot simple use the chain rule.
    I'm assuming you're ok on the product rule as you are doing inverse trigonometric functions.
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    You cannot post full solutions on here. It is against forum rules
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    (Original post by bl64)
    How would I do d2y / dx2 fpr y = (sin-1x)2?

    for dy/dx I got 2sin-1x / (1-x2)1/2

    and for d2y / dx2 I got 2-xsin-1x / (1-x2) 3/2

    the answer is supposed to be 2( (1-x2)1/2 + xsin-1x) / (1-x2) 3/2

    could someone explain how to get the correct answer

    thanks
    Can you show your quotient rule working for how you got from dy/dx to the 2nd derivative?
 
 
 
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