find equation of tangent line for x^(1/3) + y^(1/2)=2 at point (1,1)

Answer 1

Once again I can not post my answer but in the end I get
3 y = -2 x + 5

Answer 2

Can other teachers post all right ?

Answer 3

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

Answer 4

I've sent your posts to Bob and Leo. I'll send this one too.

Answer 5

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

Answer 6

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.

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Answer 7

(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 y-1 = (-3/2)(x-1)
2y - 2 = -3x + 3
2y = -3x + 5
y = (-3/2)x + 5/2

Damon, I had the same problem, now it seems to work

Answer 8

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