Question

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

Answer 1

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

Explanation:

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

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

Explanation:

`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}`.