Posted by **Anonymous** on Sunday, March 7, 2010 at 9:46pm.

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 -
**Reiny**, Sunday, March 7, 2010 at 11:13pm
√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 -
**MathMate**, Sunday, March 7, 2010 at 11:14pm
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 -
**Im**, Sunday, May 9, 2010 at 2:15am
Find dy/dx by implicit differentiation

(1-2xy^3)^5=x+4y

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