Hi! I am having trouble figuring out the derivative of ln[sin^2(x)+3]...........

Question

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 :)

Answer 1

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]

Answer 2

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