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# differentiating inverse trig functions watch

1. 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
2. (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
3. 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.
4. You cannot post full solutions on here. It is against forum rules
5. (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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