Question

show that d(tan^3x) / dx = 3sec^4x - 3sec^2x

please help,

Answer 1

d(tan^3x) / dx = 3tan^2x*sec^2x
tan^2x=sec^2x-1
(3sec^2x-3)*sec^2x=3sec^4x-3sec^2x

Answer 2

3tan^2(x) sec^2(x) = 3sec^4(x) -3sec^2 (x)
Using pythagorean identities,
3(sec^2 (x) - 1) sec^2(x) = 3sec^4(x) -3sec^2 (x)
Therefore,
3sec^4(x) - 3sec^2(x) = 3sec^4(x) -3sec^2 (x)
Got it?