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

1. anonymous

post it

2. anonymous

$f(x)=\ln(\sin^2*x)$

3. anonymous

that is sin^2(x) right?

4. anonymous

f'(x)=ln(sin^2*x)+1/2cosin1

5. anonymous

thats correct

6. anonymous

((cos(x)*2sin(x))/(sin^2(x))

7. anonymous

that 2nd equation is a bit sloppy

8. anonymous

ok thanks

9. anonymous

am i right? i'm not even sure lol

10. anonymous

yep i'm right... you can simplify it to just 2cotx

11. 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))

12. anonymous

But i am probably wrong

13. 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

14. anonymous

the derivative of ln is 1/x btw.

15. anonymous

ok but we wouldn't replace the x with whatever we are apply the natural log to? doesn't ln(x)^5 = 5lnx?