Question

Find the derivative of `tan2x` from first principles?

Answer 1

`d/dx tan2x = 2sec^2 2x`

Explanation:

By definition of the derivative `f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`
So with `f(x) = tan2x` we have;

`f'(x)=lim_(h rarr 0) ( tan(2x+2h) - tan 2x ) / h`

Using `tan A =sinA/cosA` we get

`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 `sin(A-B) -= sinAcosB - cosBsinA` we get;

`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`