calculus

posted by .

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

  • answer only - can not post derivation -

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

  • screaming frustration -

    Can other teachers post all right ?

  • Trying again -

    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

  • calculus -

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

  • Thanks -

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

  • calculus -

    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.

    =(

  • calculus -

    (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 -

    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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. math calculus

    Use implicit differentiation to find the equation of the tangent line to the curve xy^3+2xy=9 at the point (31). The equation of this tangent line can be written in the form y=mx+b where m is
  2. calculus

    5. Let f be the function given by f(x) = x3- 7x + 6. a. Find the zeros of f b. Write an equation of the line tangent to the graph of f at x = -1 c. Find the x coordinate of the point where the tangent line is parallel to the secant …
  3. COLLEGE CALCULUS.

    Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=5 at the point (4,1) . The equation of this tangent line can be written in the form y = mx+b where m is:?
  4. Calculus. I need help!

    HARDER PARTS WAS 3(x^2+y^2)^2=26(y^2+y^2) Find the equation of the tangent line to the curve (a lemniscate) 3(x^2+y^2)^2=26(y^2+y^2) at the point (-4,2). The equation of this tangent line can be written in the form y=mx+b where m is:?
  5. calculus

    To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equation of the tangent line at the point (64, 8), we know that (64, 8) is a point on the line. So we just need …
  6. Calculus

    If F(x)=x^3−7x+5, use the limit definition of the derivative to find FŒ(5), then find an equation of the tangent line to the curve y=x^3−7x+5 at the point (5, 95). FŒ(5)= The equation of the tangent line is y = x …
  7. Calculus

    1) An equation of the line contains points (7/9, 7) and (-7/9) is 2) Find the slope of the line tangent to the curve y=x^2 at the point (-0.6, 0.36) and find the corresponding equation of the tangent line
  8. AP Calculus

    Suppose that f has a continuous second derivative for all x, and that f(0)=1, f'(0)=2, and f''(0)=0. A. Does f have an inflection point at x=0?
  9. Calculus AB

    Sorry but I've got a lot of problems that I don't understand. 1) Let f(x)= (3x-1)e^x. For which value of x is the slope of the tangent line to f positive?
  10. Calculus AB

    Could someone please help me with these tangent line problems?

More Similar Questions