How do you differentiate y = (ln(x^2))^(2x+3)?
2 Answers
Note firstly that
Take the variable out of the exponent by taking logarithms:
Differentiate with the product rule (RHS) and the chain rule (LHS):
Differentiate the double log by the chain rule:
So
Thus
or, expressed in the fashion of the question
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,
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)}