Question

How do you differentiate `y = (ln(x^2))^(2x+3)`?

Answer 1

Note firstly that `y` can be expressed more simply.
`y=(ln(x^2))^(2x+3)`
`y=(2lnx)^(2x+3)`

Take the variable out of the exponent by taking logarithms:
`lny=(2x+3)ln(2lnx)`

Differentiate with the product rule (RHS) and the chain rule (LHS):
`1/ydy/dx=2ln(2lnx)+(2x+3)d/dx[ln(2lnx)]`

Differentiate the double log by the chain rule:
`d/dx[ln(2lnx)]=1/(2lnx)*d/dx[2lnx]=1/(2lnx)*2/x=1/(xlnx)`

So
`1/ydy/dx=2ln(2lnx)+(2x+3)/(xlnx)`

Thus
`dy/dx=(2lnx)^(2x+3)[2ln(2lnx)+(2x+3)/(xlnx)]`
or, expressed in the fashion of the question
`dy/dx=(ln(x^2))^(2x+3)[2lnln(x^2)+(2x+3)/(xlnx)]`

Answer 2

`dy/dx = (ln(x^2))^(2x+3){(2x+3)/(xlnx) + 2 ln 2 + 2ln(lnx)}`

Explanation:

We seek the derivative of:

`y = (ln(x^2))^(2x+3)`

We can take Natural Logarithms of both sides:

`ln y = ln {(ln(x^2))^(2x+3)}`

And using the logarithm properties, `lna^b=blna` and `lnab=lna+lnb`, this becomes:

`ln y = (2x+3)ln {ln(x^2)}`

`\ \ \ \ \ \ = (2x+3)ln {2lnx}`

`\ \ \ \ \ \ = (2x+3){ln 2 + ln(lnx)}`

Then by applying the product rule, and differentiating implicitly, we have:

`1/y \ dy/dx = (2x+3)d/dx{ln 2 + ln(lnx)} + d/dx{(2x+3)} \ {ln 2 + ln(lnx)}`

Applying the chain rule we get:

`1/y \ dy/dx = (2x+3){1/(lnx)d/dx(lnx)} + 2 {ln 2 + ln(lnx)}`

`\ \ \ \ \ \ \ \ \ \ = (2x+3)/(xlnx) + 2 ln 2 + 2ln(lnx)`

Leading to:

`dy/dx = y{(2x+3)/(xlnx) + 2 ln 2 + 2ln(lnx)}`

`\ \ \ \ \ = (ln(x^2))^(2x+3){(2x+3)/(xlnx) + 2 ln 2 + 2ln(lnx)}`