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anonymous
 5 years ago
Can someone check my Ln derivitve?
anonymous
 5 years ago
Can someone check my Ln derivitve?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\ln(\sin^2*x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is sin^2(x) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f'(x)=ln(sin^2*x)+1/2cosin1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0((cos(x)*2sin(x))/(sin^2(x))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that 2nd equation is a bit sloppy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0am i right? i'm not even sure lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep i'm right... you can simplify it to just 2cotx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well 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
 5 years ago
Best ResponseYou've already chosen the best response.0But i am probably wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why 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
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative of ln is 1/x btw.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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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