course mth 271
mr smith: here are a couple of ?s i have. thanks
4) 4x^2y-(3/y)=0y(4x^2y-(3/y))=0
4x^2y^2-3=0
deriv of 3 is zero
prod rule
(8x)(y^2)+(4x^2)(2ydy/dx)=0
8xy^2+ now here's what i dont know is it (4x^22ydy/dx) or (8x^2ydy/dx)?
(4x^2)(2ydy/dx) means (4x^2) * (2 y dy/dx), which is equal to 4 * 2 * x^2 * y * dy/dx = 8 x^2 y dy/dx.
9) (2y-x)/(y^2-3)=5
f(x)=2y-x g(x)=y^2-3
f'(x)= 2dy/dx-1 g'(x)= 2ydy/dx
quotient rule
((y^2-3)(2dy/dx-1)-(2y-x)(2ydy/dx))/(y^2-3)^2)
Very good. Excellent application of the quotient rule and the chain rule.
((2dy/dx-1)-(2y-x)(2ydy/dx))/(y^2-3)
You can't 'cancel' y^2 - 3 in one term of the numerator with a y^2 - 3 in the denominator. Unless the y^2 - 3 is a factor of both terms in the numerator, this won't work, because division is just multiplication by a reciprocal and hence distributes over addition or subtraction.
In any case this expression gets complicated.
You would be better off to avoid the quotient rule by multiplying both sides by the denominator y^2 - 3:
The equation rearranges to 2y - x = 5 ( y^2 - 5), so that
5 y^2 - 2 y + x - 5 = 0. Implicitly differentiating this expression gives us
10 y dy/dx - 2 dy/dx + 1 = 0 so that
(10 y - 2) dy/dx = -1 and
dy/dx = -1 / (10 y - 2).
im not sure where to go from here. the book says the answer is -1/(10y-2)"
See my notes and let me know if you have questions. You appear to be doing very well.