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Expert: - 5/4/2010


Question
Could you tell me if my answers to these questions are correct or if they are not correct maybe give me a little help this would be appreciated

1. Calculate dy/dx whan y^2 = (3x + 1)^3/9x + 2
My answer   -3(3x + 1)^2/y^2(9x + 2)^2
2. If y = 3x^2 + 6x and x = 2t^2 + t find dy/dx
My Answer 9x^2 + 6/4t + 1
Thanks

Answer
1. If y² = (3x+1)³/9x + 2, that it is y² = f/g.
Taking the derivative of both sides gives 2yy' = (gf' - fg')/g².

If y = 3x² + 6x and x = 2t² + t, then dy/dx is not concerned with t.  It is 6x + 6.
To do the other sides, take f = (3x+1)³, so f' = 3(3x+1)², and g(x) = 9x, so g'(x) = 9.
It looks like the 2 is on the end by itself, so the derivative of constants is 0.

However, it seems like maybe that want dy/dt, and that is just like fraction.
It is dy/dt = (dy/dx)(dx/dt).
It is known that dy/dx = 6x + 6 and dx/dt = 4t + 1.

To get the answer in terms of t, put in what x(t) is for x.

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