Question

Can someone check my Ln derivitve?

Answer 1

post it

Answer 2

\[f(x)=\ln(\sin^2*x)\]

Answer 3

that is sin^2(x) right?

Answer 4

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

Answer 5

thats correct

Answer 6

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

Answer 7

that 2nd equation is a bit sloppy

Answer 8

ok thanks

Answer 9

am i right? i'm not even sure lol

Answer 10

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

Answer 11

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))

Answer 12

But i am probably wrong

Answer 13

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

Answer 14

the derivative of ln is 1/x btw.

Answer 15

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