How do you find the derivative of #y=sqrt( x+sqrt( x+sqrt( x)))#?

Question

4519 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T15:34:26

#dy/dx=1/{2y}({2\sqrtx+1}/{4\sqrtx(y^2-x)}+1)#

Given that

#y=\sqrt{x+\sqrt{x+\sqrtx}}#

#y^2=x+\sqrt{x+\sqrt{x}}#

#y^2-x=\sqrt{x+\sqrt{x}}#

#(y^2-x)^2=x+\sqrt{x}#

Differentiating above equation w.r.t. #x# using chain rule as follows

#d/dx(y^2-x)^2=d/dx(x+\sqrtx)#

#2(y^2-x)d/dx(y^2-x)=1+1/{2\sqrtx}#

#2(y^2-x)(2ydy/dx-1)=1+1/{2\sqrtx}#

#2ydy/dx-1={2\sqrtx+1}/{4\sqrtx(y^2-x)}#

#dy/dx=1/{2y}({2\sqrtx+1}/{4\sqrtx(y^2-x)}+1)#