If #xsqrt(1 + y) + ysqrt(1 + x) = 0#. Then how will you prove that #dy/dx = -1/(1 + x)^2#??

Question

71376 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-24T17:23:31

#dy/dx=-1/(x+1)^2#

#xsqrt(1 + y) + ysqrt(1 + x) = 0#

#xsqrt(1 + y)=-ysqrt(1 + x)#

#x^2*(1 + y)=(-y)^2*(1 + x)#

#x^2*(1 + y)=y^2*(1 + x)#

#x^2 + x^2*y=y^2+y^2*x#

#x^2 -y^2=y^2*x-x^2*y#

#(x+y) * (x-y) = -xy * (x-y)#

#x+y = -xy#

#x = -xy-y#

#x = -y* (x+1)#

#y=-x/(x+1)#

#dy/dx=[-1*(x+1)-(-x)*1]/(x+1)^2#

#dy/dx=-1/(x+1)^2#

1) I solved this equation for y.

2) I differentiated both sides.