Basic formula for finding the ln of a fxn:
ln u = 1/u dx (u)
So in this case U = e^2x/x^2-4 ^3
So: 1/e^2x/blah blah blah then get the derivative of e^2x/blahblah.
So basically 1/u dx (e^blahblah)
Find the derivative of the following:
f(x)=ln(e^(2x)/(x^2-4)^3
Please show all steps so I can follow
Thanks!
3 answers
You have unbalanced parens, but I'll take a stab at it.
f = u/v, so
f' = (u'v-uv')/v^2
u=ln(e^(2x))
u' = 1/e^(2x) * 2e^2x = 2
note: ln(e^(2x)) = 2x (definition of log)
v = (x^2-4)^3
v' = 3(x^2-4)^2 (2x) = 6x(x^2-4)^2
f' = [(2)(x^2-4)^3 - (2x)(6x)(x^2-4)^2]/(x^2-4)^6
= 2(x^2-4)^2 ((x^2-4)-6x^2) / (x^2-4)^6
= -2(5x^2+4)/(x^2-4)^4
f = u/v, so
f' = (u'v-uv')/v^2
u=ln(e^(2x))
u' = 1/e^(2x) * 2e^2x = 2
note: ln(e^(2x)) = 2x (definition of log)
v = (x^2-4)^3
v' = 3(x^2-4)^2 (2x) = 6x(x^2-4)^2
f' = [(2)(x^2-4)^3 - (2x)(6x)(x^2-4)^2]/(x^2-4)^6
= 2(x^2-4)^2 ((x^2-4)-6x^2) / (x^2-4)^6
= -2(5x^2+4)/(x^2-4)^4
we could do some preliminary simplification
reading it as:
f(x)=ln (e^(2x)/(x^2-4)^3 )
= ln e^(2x) - ln (x^2-4)^3
= 2x - 3ln(x^2 - 4)
f ' (x) = 2 - 3(2x)/(x^2 - 4)
= 2 - 6x/(x^2 - 4) = (2x^2 - 6x - 8)/(x^2-4)
notice that is is a different interpretation as Steve's, since your brackets are mis-matched, who can tell.
reading it as:
f(x)=ln (e^(2x)/(x^2-4)^3 )
= ln e^(2x) - ln (x^2-4)^3
= 2x - 3ln(x^2 - 4)
f ' (x) = 2 - 3(2x)/(x^2 - 4)
= 2 - 6x/(x^2 - 4) = (2x^2 - 6x - 8)/(x^2-4)
notice that is is a different interpretation as Steve's, since your brackets are mis-matched, who can tell.
1 answer