Question

How do you differentiate `f(x)= x^2/ (lnx)` twice using the quotient rule?

Answer 1

`f''(x)=(2ln^2x-3lnx+2)/ln^3x`

Explanation:

The quotient rule states that

`d/dx(g(x)/(h(x)))=(h(x)g'(x)-g(x)h'(x))/(h(x))^2`

So, for `f(x)`, we see that

`g(x)=x^2`
`h(x)=lnx`

Differentiate both of these:

`g'(x)=2x`
`h'(x)=1/x`

Plugging these both in, we see that

`f'(x)=(lnx(2x)-x^2(1/x))/(ln^2x`

Which simplifies to be

`f'(x)=(2xlnx-x)/ln^2x`

We can apply the quotient rule to this again to find the second derivative. We will redefine our functions being divided:

`g(x)=2xlnx-x`
`h(x)=ln^2x`

However, this time, finding `g'(x)` and `h'(x)` will not be so straightforward.

The `2xlnx` term of `g(x)` will require the product rule to differentiate.

We see that

`g'(x)=lnxd/dx(2x)+2xd/dx(lnx)-1`

`g'(x)=lnx(2)+2x(1/x)-1`

`g'(x)=2lnx+1`

Finding `h'(x)` will require use of the chain rule:

`h'(x)=2lnxd/dx(lnx)=2lnx(1/x)=(2lnx)/x`

Thus, we see that

`g(x)=2xlnx-x`
`h(x)=ln^2x`

`g'(x)=2lnx+1`
`h'(x)=(2lnx)/x`

Hence the second derivative equals

`f''(x)=(ln^2x(2lnx+1)-(2xlnx-x)(2lnx)/x)/ln^4x`

Expanding completely, this equals

`f''(x)=(2ln^3x+ln^2x-4ln^2x+2lnx)/ln^4x`

Combining the `ln^2x` terms and dividing a common `lnx` term from the fraction, we see that

`f''(x)=(2ln^2x-3lnx+2)/ln^3x`