How do you use implicit differentiation to find dy/dx given #sqrtx+sqrt(yx^2)=x#?

Question

Luke Alvarez · Accepted Answer

#dy/dx = (2sqrty)/x-sqrty/(xsqrtx)-(2y)/x# Differentiate both sides of the equation with respect to #x#: #d/dx( sqrtx + sqrt (yx^2) )= d/dx (x)# #d/dx( sqrtx) + d/dx( sqrt (yx^2) )= 1# #1/(2sqrtx) + 1/(2sqrt(yx^2)) d/dx (yx^2) = 1# #1/(2sqrtx) + 1/(2sqrt(yx^2)) (2xy+x^2dy/dx) = 1# #1/(2sqrtx) + (2xy)/(2sqrt(yx^2)) +x^2/(2sqrt(yx^2))dy/dx = 1# Note that both #x > 0# and #y > 0# if the equation is in the real domain, otherwise #sqrt x# and #sqrt(yx^2)# would not be defined, so #x/sqrt(x^2) = 1# and #y/sqrty = sqrty# #1/(2sqrtx) +sqrty +x/(2sqrt(y))dy/dx = 1# #x/(2sqrt(y))dy/dx = 1-1/(2sqrtx) -sqrty# #dy/dx = (2sqrty)/x(1-1/(2sqrtx) -sqrty)# #dy/dx = (2sqrty)/x-sqrty/(xsqrtx)-(2y)/x#

Cameron Alexander · Answer

undefined To use implicit differentiation to find dy/dx given sqrt(x) + sqrt(yx^2) = x, follow these steps:

1. Differentiate both sides of the equation with respect to x.
2. Apply the chain rule and the power rule as needed.
3. Solve the resulting equation for dy/dx.

Differentiating both sides with respect to x:

d/dx(sqrt(x) + sqrt(yx^2)) = d/dx(x)

Apply the chain rule and the power rule:

(1/2)*(1/sqrt(x))*(1) + (1/2)*(1/sqrt(yx^2))*(2x(dy/dx)) = 1

Simplify and solve for dy/dx:

(1/2)*(1/sqrt(x)) + (x/y)*(1/sqrt(yx^2))(dy/dx) = 1

dy/dx = [1 - (1/2)*(1/sqrt(x))] / [(1/2)*(x/y)*(1/sqrt(yx^2))]