有兩題數唔識做

1, y = [ln(x)][sin(e^3x)]
dy/dx=d[ln(x)]/dx *[sin(e^3x)] + [ln(x)]*d[sin(e^3x)]/dx
dy/dx=1/x *[sin(e^3x)] + [ln(x)]*cos(e^3x)*d(e^3x)/dx
dy/dx=1/x *[sin(e^3x)] + 3[ln(x)]*cos(e^3x)*(e^3x)

2. show that if y = tan^-1(x); then dy/dx = 1/1+x^2

y = tan^-1(x)
tany = x
d(tany)/dx = dx/dx
sec^2(y)* dy/dx = 1
dy/dx=1/sec^2(y)-----------------(1)

to find sec^2(y) in terms of x:
tany = x
(tany)^2 = x^2
(tany)^2 + 1 = x^2 + 1
(secy)^2 = x^2 + 1
Put back into (1)
dy/dx=1/(secy)^2 = 1/x^2 + 1

I only a P.5.

第一題你想點
求derivative?
y=[ln(x)]{[sin(e^(3x)]}
Let k=e^(3x),m=sin k
dm/dx
=(dm/dk)(dk/dx)
=3 cos [e^(3x)] e^(3x)
dy/dx=[ln(x)]{3 cos [e^(3x)] e^(3x)}+{[sin(e^(3x)]}/x
=3 cos [e^(3x)] ln (x)+{sin[e^(3x)]}/x

(2)
Let me think a while, I will answer you in few days.