Tuesday

January 17, 2017
Posted by **jack** on Sunday, June 16, 2013 at 7:32pm.

- answer only - can not post derivation -
**Damon**, Sunday, June 16, 2013 at 7:53pmOnce again I can not post my answer but in the end I get

3 y = -2 x + 5 - screaming frustration -
**Damon**, Sunday, June 16, 2013 at 7:54pmCan other teachers post all right ?

- Trying again -
**Damon**, Sunday, June 16, 2013 at 7:56pmno 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 - calculus -
**Ms. Sue**, Sunday, June 16, 2013 at 7:58pmI've sent your posts to Bob and Leo. I'll send this one too.

- Thanks -
**Damon**, Sunday, June 16, 2013 at 7:59pmThanks - I can not really answer this because it will not let me write out derivative method.

- calculus -
**Writeacher**, Sunday, June 16, 2013 at 8:00pmBob 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.

=( - calculus -
**Reiny**, Sunday, June 16, 2013 at 9:27pm(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 - calculus -
**Damon**, Monday, June 17, 2013 at 9:39amtangent 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