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July 26, 2014

July 26, 2014

Posted by **KIKI** on Tuesday, August 2, 2011 at 10:47am.

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 11:09amLet 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 3:55pmthank 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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