Posted by KIKI on Tuesday, August 2, 2011 at 10:47am.
Hi! I am having trouble figuring out the derivative of ln[sin^2(x)+3]...........
If it weren't for the "3" I know I could just "move" the exponent of 2 from sin^2 to in front of the ln,....but because of the 3 I can't do this....So what should I do?
Thank you very much for your help! Have a nice day :)

Calculus  drwls, Tuesday, August 2, 2011 at 11:09am
Let u = sin^2x + 3
f(x) = ln[sin^2(x)+3]
= ln u(x)
df/dx = df/du*du/dx
= (1/u)*2sinx*cosx
= sin(2x)/[sin^2x+3]

Calculus  KIKI, Tuesday, August 2, 2011 at 3:55pm
thank you! but how did you go from......(1/u)*2sinx*cosx....to......sin(2x)/[sin^2x+3].............? i was just confused by this..thanks
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