another derivation thats got me stumped!!?

Question

what is f '(x) of
f(x) = sin^3(x) - cos^3(x)

another derivation thats got me puzzled

Answer 1

f'(x) = 3sin^2(x)*cosx + 3cos^2(x)*sinx

Answer 2

f(x) = sin^3(x) - cos^3(x)

We can either solve this as-is, or use some nifty methods to change that and then differentiate. Let's solve it as-is. This is the same as

f(x) = [sin(x)]^3 - [cos(x)]^3

Differentiate using the power rule and chain rule.

f'(x) = 3[sin(x)]^2 cos(x) - 3[cos(x)]^2 (-sin(x))

Which simplifies as

f'(x) = 3sin^2(x) cos(x) + 3cos^2(x) sin(x)

Factor,

f'(x) = 3sin(x)cos(x) [ sin(x) + cos(x) ]

Answer 3

Write as:-
f(x) = (sin x)³ - (cos x)³
f ` (x) = 3(sin² x).cos x - 3.(cos² x).(- sin x)
f ` (x) = 3 sin² x.cos x + 3sin x.cos² x
f ` (x) = 3.sin x.cos x.(sin x + cos x)

Answer 4

f(x)=sin^3(x)-cos^3(x)
therefore
f'(x)=3sin^2(x)cos(x)-3cos^2(x)(-sin(x))
=3sin^2(x)cos(x)+3cos^2(x)sin(x)