dy/dx (ln(3x))^2?

Question

I'm forgetting how to differentiate!

I have the correct answer (ti-89): I was hoping someone could show me *how it's done. (I should have explained better)

thanks mrgets

Answer 1

Hello,

2ln(3x)*(1/3x)*3
[2ln(3x)]/x

Hope This Helps!

Answer 2

Solution #1 (nota : dy/dx = y')

y = (ln(3x))^2

ln(y) = 2ln(ln(3x))

y'/y = (2)' ln(ln(3x)) + (ln(ln(3x)))' (2)

= 0 + (1/ln(3x)) (1/(3x))(3) (2)

y' = (ln(3x))^2 2/x(ln(3x)

y' = 2/x(ln(3x))



Solution #2

y = (ln(3x))^2

y = (ln(3x)) (ln(3x))

y' = (ln(3x))' (ln(3x)) + (ln(3x))' (ln(3x))

y' = (1/3x)(3) ln(3x) + (1/3x)(3) ln(3x)

y' = (1/x) ln(3x) + (1/x) ln(3x)

y' = (2/x)ln(3x)


Solution #3 (simplest solution)

y = (ln(3x))^2

y' = 2 ln(3x) (1/3x)(3)

y' = (2/x)ln(3x)

Answer 3

2ln(3x)/x