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