the question is
y = (sin 2x)^3 (x^4 - 4x)^6 / (cos x) + e^(3x), find y' using log diff.
what i did so far is,
ln y = ln[(sin 2x)^3] + ln[(x^4 - 4x)^6] - @@@@@@@@@@
at @@@@@@
should it be ln[(cos x) + e^(3x)] or ln(cos x) - ln [e^(3x)] ?
the question is
y = (sin 2x)^3 (x^4 - 4x)^6 / (cos x) + e^(3x), find y' using log diff.
what i did so far is,
ln y = ln[(sin 2x)^3] + ln[(x^4 - 4x)^6] - @@@@@@@@@@
at @@@@@@
should it be ln[(cos x) + e^(3x)] or ln(cos x) - ln [e^(3x)] ?
i think i would go with ln[(cos x) + e^(3x)]
y = (sin 2x)^3 (x^4 - 4x)^6 / (cos x) + e^(3x)
ln y = ln[(sin 2x)^3] + ln[(x^4 - 4x)^6] - ln[(cos x) + e^(3x)]
= 3ln(sin 2x) + 6ln(x^4 - 4x) - ln[(cos x) + e^(3x)]
(1/y)(y') = (3 cos 2x/sin 2x) + (6(4x^3 - 4)/x^4 - 4x) - [(-sin x + 3e^3x)/(cos x) + e^(3x)]
y' = [y][(3 cos 2x/sin 2x) + (6(4x^3 - 4)/x^4 - 4x) - [(-sin x + 3e^3x)/(cos x) + e^(3x)]]
y' = [(sin 2x)^3 (x^4 - 4x)^6 / (cos x) + e^(3x)][(3 cos 2x/sin 2x) + (6(4x^3 - 4)/x^4 - 4x) - [(-sin x + 3e^3x)/(cos x) + e^(3x)]]
did i do it right...?
would be highly appreciate if someone point me out if i did something wrong
thanks