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A particle moves along the curve f(x). As it reaches the point (3; 2), the y-coordinate is increasing by 1.2 cm/sec. At what rate is the x-coordinate changing at that instant?
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y=\[\sqrt[3]{x^{2}-1}\]
here they actually give you the formula
I take the derivative.
but then what?
yes, don't forget that \(x\) is a function of \(t\) also, your answer should have a \(x'\) in it
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oh, so I just plug in 1.2 to x' and 3 into x and get y'?
at the end, yes you should get as a first step \[y'=\frac{2xx'}{3\sqrt{x^2-1}^2}\]
I got 0.6
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