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If f(x) = 3 sin x + ln(4x), find f '(x). f '(x) =
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f(x) = 3 sin x + ln(4x) f'(x) = d/dx (3 sin x) + d/dx (ln(4x)) d/dx sinx = cosx d/dx lnx = 1/x Can you do it now?
3cosx+....
:(
d/dx ln(ax) = (1/ax) (a) = ...
3cosx+1/x right??
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It should be!
I meant it is correct.
thanks
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