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

2 Answers
Jun 14, 2018

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)]

Jun 14, 2018

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)}