calculus
posted by jack on .
find equation of tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)

Once again I can not post my answer but in the end I get
3 y = 2 x + 5 
Can other teachers post all right ?

no luck  hope someone can help you.
the idea is to do f(d+dx)  f(x)
where f(x+dx) = f(x) + (dy/dx) dx 
I've sent your posts to Bob and Leo. I'll send this one too.

Thanks  I can not really answer this because it will not let me write out derivative method.

Bob knows. He has cleaned out the banned list, and said he'd check more specifically later.
I've texted Leo ... I'm surprised there's been no answer.
=( 
(1/3)x^(2/3) + (1/2)y^(1/2) dy/dx = 0
at (1,1)
(1/3)(1) + (1/2)(1)dy/dx =0
dy/dx =(1/2) / (1/3) = 3/2
so y1 = (3/2)(x1)
2y  2 = 3x + 3
2y = 3x + 5
y = (3/2)x + 5/2
Damon, I had the same problem, now it seems to work 
tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)

(1/3) / x^(2/3) + (1/2) dy/dx /y^(1/2)=0
so
(1/2)dy/dx = (1/3) y^(1/2)/x^(2/3)
at (1.1)
dy/dx = (2/3)
then
y = 2x/3 + b
1 = 2/3 + b
b = 5/3
so
y = 2x/3 + 5/3
3 y = 2 x + 5