Question

How do we find the derivative of `tanx` from first principal?

Answer 1

`d/(dx)tanx=sec^2x`

Explanation:

According to first principal, if `y=f(x)`, then

`(dy)/(dx)=Lt_(deltax->0)(f(x+deltax)-f(x))/(deltax)`

Here we have `y=f(x)=tanx`

hence `f(x+deltax)=tan(x+deltax)`

and `(dy)/(dx)=Lt_(deltax->0)(tan(x+deltax)-tanx)/(deltax)`

`=Lt_(deltax->0)(sin(x+deltax)/cos(x+deltax)-sinx/cosx)/(deltax)`

`=Lt_(deltax->0)(sin(x+deltax)cosx-cos(x+deltax)sinx)/(cosxcos(x+deltax)deltax)`

`=Lt_(deltax->0)(sin(x+deltax-x))/(cosxcos(x+deltax)deltax)`

`=Lt_(deltax->0)(sin(deltax)/(deltax) xx1/(cosxcos(x+deltax)))`

`=1xxsec^2x=sec^2x`