#1 is good
in #2 you could go further before taking any derivative
y'(1/y) = ln(sin^2xtan^4x) - ln(x^2+1)^2 or
y'(1/y) = ln(sin^2x) + ln(tan^4x) - 2ln(x^2 + 1)
y'(1/y) = 2ln(sinx) + 4ln(tanx) - 2ln(x^2+1)
and now differentiate it
In #3 from your second last line of
y' = (1/x)^2 + ln x*(-1/x^2)y I would do
y ' = y(1 - lnx)/x^2 and leave it like that
#4 is exponential differentiation.
Notice the base is a constant, not a variable
recall if y = a^u
then y' = (ln a)(a^u)(u')
so for y = 2^(-x^3)
y' = ln2(-3x^2)(2^(-x^3))
If you want to do it by the same method as the above, you would get
lny = ln(2^(-x^3))
now take the derivative
y'/y = -ln2(3)x^2
y' = -3ln2(x^2)y
= my answer above
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