Chain rule! CALC 2! y=sin(cos(e^(tan(3x))))

y=sin(cos(e^(tan(3x))))
y`=
please help and explain where i start

y' = d/dx sin(cos(e^(tan(3x)))) ... [outer function sin(x), inner function cos(e^(tan(3x)))]
= cos(cos(e^(tan(3x)))) * d/dx cos(e^(tan(3x))) ... [outer: cos, inner: e^(tan(3x))]
= cos(cos(e^(tan(3x)))) * -sin(e^(tan(3x))) * d/dx e^(tan(3x)) ... [outer: e^x, inner: tan(3x)]
= cos(cos(e^(tan(3x)))) * -sin(e^(tan(3x))) * e^(tan(3x)) * d/dx tan(3x) ... [outer: tan, inner: 3x]
= cos(cos(e^(tan(3x)))) * -sin(e^(tan(3x))) * e^(tan(3x)) * sec^2(3x) * d/dx 3x
= cos(cos(e^(tan(3x)))) * -sin(e^(tan(3x))) * e^(tan(3x)) * sec^2(3x) * 3
= -3cos(cos(e^(tan(3x))))sin(e^(tan(3x)))e…

Just peel the layers off one at a time with chain rule.

I understand this answerer showed more work hence it make senses he or she gets best answer; however, I fail to see the reason for the two thumbs down. My answer was essentially the same, I felt no need to show the extra d/dx each step of the way, and did it one going, seeing as that is efficient.

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Work your way in:

y = sin(cos(e^(tan(3x))))
y' = cos(cos(e^(tan(3x)))) * -sin(e^(tan(3x))) * e^(tan(3x)) * sec²(3x) * 3
y' = -3e^(tan(3x)) * sec²(3x) * sin(e^(tan(3x))) * cos(cos(e^(tan(3x))))