Post a New Question

calculus

posted by .

Point P(a,b) is on the curve
square root of x + square root of y =1.

Show that the slope of the tangent at P is : - square root of (b/a).

  • calculus -

    √x + √y = 1 or
    x^(1/2) + y^(1/2) = 1
    differentiate implicitly
    (1/2)x^(-1/2) + (1/2)y^(-1/2)dy/dx = 0
    dy/dx = -x^(-1/2)/y^(-1/2)
    = - y^(1/2)/x^(1/2)
    = - √y/√x
    = - √(y/x)

    so at the point (a,b)
    dy/dx = - √(b/a)

  • implicit differentiation -

    Given
    √x + √y = 1
    Apply implicit differentiation:
    1/(2√x) + 1/(2√y)*(dy/dx) = 0
    Transposing and solving for dy/dx:

    dy/dx = -√(y/x)

  • calculus -

    Find dy/dx by implicit differentiation
    (1-2xy^3)^5=x+4y

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question