Find the derivative of tan2x from first principles?
1 Answer
d/dx tan2x = 2sec^2 2x
Explanation:
By definition of the derivative
So with
f'(x)=lim_(h rarr 0) ( tan(2x+2h) - tan 2x ) / h
Using
f'(x) = lim_(h rarr 0) ( sin(2x+2h)/cos(2x+2h) - (sin2x)/(cos2x) ) / h
" " = lim_(h rarr 0) 1/h{ sin(2x+2h)/cos(2x+2h) - (sin2x)/(cos2x) }
" " = lim_(h rarr 0) 1/h{ (sin(2x+2h)cos2x - cos(2x+2h)sin2x)/(cos(2x+2h)cos2x) }
Using the identity
f'(x) = lim_(h rarr 0) 1/h{ (sin(2x+2h-2x))/(cos(2x+2h)cos2x) }
" " = lim_(h rarr 0) 1/h{ (sin(2h))/(cos(2x+2h)cos2x) }
" " = lim_(h rarr 0) 1/(cos(2x+2h)cos2x) * (sin2h)/h
" " = lim_(h rarr 0) 2/(cos(2x+2h)cos2x) * (sin2h)/(2h)
" " = lim_(h rarr 0) 2/(cos(2x+2h)cos2x) * lim_(h rarr 0) sin(2h)/(2h)
We now have to rely on some standard limits:
lim_(h rarr 0)sin h/h =1 => lim_(h rarr 0) (sin 2h)/(2h) =1
And so we get:
f'(x) = 2/(cos2xcos2x) * 1
" " = 2/(cos^2 2x)
" " = 2sec^2 2x
