Question

How do you use the chain rule to differentiate `sqrt(ln4x)`?

Answer 1

`1/(2xsqrt(ln4x))`

Explanation:

Apply the chain rule:
`d/dxsqrt(ln4x)`

So the chain rule states that `(df(u))/dx=(df)/(du)*(du)/dx`.
Here, `f=sqrt(u)`, and `u=ln4x`.

According to the power rule `d/dxx^a=ax^(a-1)`.

So `d/(du)sqrt(u)`
`=d/(du)u^(1/2)`
`=1/2(1/sqrt(u))`
`=1/(2sqrt(u))`

Now that we've solved `(df)/(du)`, we have to solve `(du)/dx`.

But we can't. We cannot differentiate `ln4x` directly.

Use the chain rule again: Here, `f=lnu`, and `u=4x`
We know that `d/(du)lnu=1/u`, and that `d/dx4x=4`

So now we have `d/(dx)ln4x=1/u*4`

But `u=4x,` so input:

`1/(4x)*4=1/x`

So now we have `1/(2sqrt(u))*1/x`

`=1/(2sqrt(ln4x))*1/x`

`=1/(2xsqrt(ln4x))`