How do you use the chain rule to differentiate #y=sqrt(1/(x+1))#?

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Answer 1 · 2018-07-22T13:48:21

#dy/dx=-1/{2(x+1)^{3/2}}#

Given function:

#y=\sqrt{1/{x+1}}#

#y=1/{(x+1)^{1/2}}#

#y=(x+1)^{-1/2}#

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

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

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

#=-1/2(x+1)^{-3/2}(1)#

#=-1/{2(x+1)^{3/2}}#

Answer 2 · 2018-07-22T14:05:12

# dy/dx=-1/{2(x+1)sqrt(x+1)}=-1/{2(x+1)^(3/2)}=-1/{2(sqrt(x+1))^3}#.

#y=sqrt{1/(x+1)}=1/sqrt(x+1)#.

Let, #(x+1)=u, sqrtu=sqrt(x+1)=v#.

Thus, #y=1/v, v=sqrtu, u=x+1#.

Thus, #y# is a function of #v, v" of "u, &, u" of "x#.

Therefore, by the Chain Rule,

#dy/dx=dy/(dv)(dv)/(du)(du)/dx..........(ast)#.

Now, #dy/(dv)=d/(dv){1/v}=-1/v^2......(ast^1)#,

#(dv)/(du)=d/(du){sqrtu}=1/(2sqrtu)........(ast^2)#, and,

#(du)/dx=d/dx{x+1}=1................................(ast^3)#.

We combine #(ast^1),(ast^2),(ast^3) and (ast)#, to get,

#dy/dx=(-1/v^2)(1/(2sqrtu))(1),#

#=(-1/u)(1/(2sqrt(x+1)))#.

#:.dy/dx=-1/{2(x+1)sqrt(x+1)}=-1/{2(x+1)^(3/2)}=-1/{2(sqrt(x+1))^3}#.