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

1 Answer
Aug 4, 2017

Answer:

#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#