anonymous
  • anonymous
Can someone check my Ln derivitve?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
post it
anonymous
  • anonymous
\[f(x)=\ln(\sin^2*x)\]
anonymous
  • anonymous
that is sin^2(x) right?

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anonymous
  • anonymous
f'(x)=ln(sin^2*x)+1/2cosin1
anonymous
  • anonymous
thats correct
anonymous
  • anonymous
((cos(x)*2sin(x))/(sin^2(x))
anonymous
  • anonymous
that 2nd equation is a bit sloppy
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
am i right? i'm not even sure lol
anonymous
  • anonymous
yep i'm right... you can simplify it to just 2cotx
anonymous
  • anonymous
well i am thinking that because of the properties of natural log we use the coefficent of the ln for the numerator and whatever follows the ln as the denomator. so becaues we just have ln(sin^2x) we then get ((1)/2cos(x))
anonymous
  • anonymous
But i am probably wrong
anonymous
  • anonymous
why would you do that though? I guess you can simplify it by using 2ln(sin(x)), and then just move the 2 out. which would get the same answer..., because ln (sinx) = cos(x)/sin(x) = cot(x) *2
anonymous
  • anonymous
the derivative of ln is 1/x btw.
anonymous
  • anonymous
ok but we wouldn't replace the x with whatever we are apply the natural log to? doesn't ln(x)^5 = 5lnx?

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