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

1 Answer
Shwetank Mauria
Aug 4, 2017

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

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