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Find the derivative of y y=sin^-1 (3/t^2)
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i suggest finding the derivative implicitly if you don't remember derivative for inverse sine function
What I got up to so far: y'= -6t^-3/ cos y I dont know where to proceed from there. :/
\[\sin(y)=\frac{3}{t^2} \\ \text{ or } \sin(y)=3t^{-2}\]
you are done unless you want to write y' in terms of t
only
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you can do this by making a reference right triangle for the equation sin(y)=3/t^2
|dw:1448246998396:dw| find adj and then find cos(y)
okay, I got cos y= sqr (t^4-9) / t^2, so do i plug that into y' equals thingy?
you replace cos(y) with that yes
I got the same answer as the one in my textbook after doing that. Thanks! Now I know how to do this lol
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